update: 添加问题“1266.访问所有点的最小时间”的代码和题解 (#1321)

LetMeFly666 · web-flow · commit 8f477720f657 · 2026-01-12T23:54:12.000+08:00
* archive: before.换设备 * 1266: AC.cpp+py+java+go+rust (#1321) (#1318) + word(en) cpp - AC,100.00%,16.04% py - AC,100.00%,5.26% java - AC,91.80%,8.20% go - AC,100.00%,85.71% rust - AC,100.00%,75.00% * update: 添加问题“1266.访问所有点的最小时间”的代码和题解 (#1321) Signed-off-by: LetMeFly666 <Tisfy@qq.com> --------- Signed-off-by: LetMeFly666 <Tisfy@qq.com>
diff --git a/.commitTitleExtra b/.commitTitleExtra
diff --git a/.commitmsg b/.commitmsg
diff --git a/.github/workflows/deployHexo.yml b/.github/workflows/deployHexo.yml
@@ -36,7 +36,7 @@ jobs:
         uses: actions/checkout@v3
         with:
           persist-credentials: true
-          fetch-depth: 1
+          fetch-depth: 2
       
       - name: Whether Skip Deploy(.nodeploy|filePath)
         id: deploy-check
@@ -78,6 +78,7 @@ jobs:
       - name: Fetch and checkout each branch (shallow)
         if: env.SKIP_DEPLOY != 'true'
         run: |
+          # TODO: Let's change this(only needed branch)
           # git clone git@github.com:LetMeFly666/LeetCode.git --depth=1
           # cd LeetCode
           # git rev-list --count HEAD
@@ -169,51 +170,56 @@ jobs:
         run: |
           set -e
 
-          json_file="../website_modifyTime/LastModifiedTime.json"
+          # git rev-list --count HEAD
+          json_file_name="LastModifiedTime.json"
+          json_file="../website_modifyTime/$json_file_name"
 
-          # 获取本次 push 里 master 分支变动的文件列表
-          # TODO: Let's处理删除removed
-          # TODO: Let's solve this
-          echo '${{ toJSON(github.event) }}' | jq .
-          changed_files=$(jq -r '
-            [
-              .commits[].added[]?,
-              .commits[].modified[]?
-            ]
-            | unique
-            | .[]
-          ' "$GITHUB_EVENT_PATH")
-          echo "Changed files:"
-          echo "$changed_files"
+          if [ -f "$json_file" ]; then
+            json_content=$(cat "$json_file")
+          else
+            json_content='{}'
+            echo "original Modified Json File Not exists"
+          fi
 
-          # 保存原 json 内容顺序
-          tmp_json=$(mktemp)
-          cp "$json_file" "$tmp_json"
+          commit_time=$(git show -s --format=%cI HEAD)
+          echo "Commit time: $commit_time"
 
-          # 遍历每个文件，看是否是 Solutions 下的 md 文件
-          updated=false
-          commit_time=$(git show -s --format=%cI ${{ github.sha }})  # ISO 8601 commit 时间
+          has_modified=false
+          while read -r status file; do
+            [[ "$file" == Solutions/*.md ]] || continue
 
-          while IFS= read -r file; do
-            # 只处理 Solutions/*.md 文件
-            if [[ "$file" == Solutions/*.md ]]; then
-              filename=$(basename "$file")
-              # 使用 jq 更新 JSON 值
-              # 保持原有顺序，用临时文件写回
-              jq --arg key "$filename" --arg value "$commit_time" '(.[$key] = $value) | .' "$tmp_json" > "$tmp_json.tmp" && mv "$tmp_json.tmp" "$tmp_json"
-              updated=true
-            fi
-          done <<< "$changed_files"
+            filename=$(basename "$file")
+            has_modified=true
+
+            case "$status" in
+              A|M)
+                echo "Updated $filename"
+                json_content=$(echo "$json_content" | jq \
+                  --arg k "$filename" \
+                  --arg v "$commit_time" \
+                  '. + {($k): $v}')
+                ;;
+              D)
+                echo "Deleted $filename"
+                json_content=$(echo "$json_content" | jq \
+                  --arg k "$filename" \
+                  'del(.[$k])')
+                ;;
+            esac
+          done < <(git diff --name-status HEAD~1 HEAD)
 
-          if [ "$updated" != true ]; then
-            echo "No Solutions/*.md file changed, skip JSON update."
+          if [ "$has_modified" = false ]; then
+            echo "no modified Solutions/*.md files"
             exit 0
           fi
 
+          # 按 key 字典序排序并写回
+          jq -S --indent 4 '.' <<<"$json_content" > "$json_file"
+          echo "Updated $json_file"
+
           # 提交修改
-          cp "$tmp_json" "$json_file"
           cd ../website_modifyTime
-          git add "$json_file"
+          git add "$json_file_name"
           TOCOMMIT=$(printf "%s\n\n%s" "$MESSAGE" "$HASH")
           git commit --allow-empty -m "$TOCOMMIT"
           git push
diff --git a/Codes/1266-minimum-time-visiting-all-points.cpp b/Codes/1266-minimum-time-visiting-all-points.cpp
@@ -0,0 +1,20 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-01-12 23:24:06
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-01-12 23:28:12
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
+        int ans = 0;
+        for (int i = 1; i < points.size(); i++) {
+            ans += max(abs(points[i][0] - points[i - 1][0]), abs(points[i][1] - points[i - 1][1]));
+        }
+        return ans;
+    }
+};
diff --git a/Codes/1266-minimum-time-visiting-all-points.go b/Codes/1266-minimum-time-visiting-all-points.go
@@ -0,0 +1,21 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-01-12 23:24:06
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-01-12 23:35:23
+ */
+package main
+
+func abs1266(a int) int {
+    if a < 0 {
+        return -a
+    }
+    return a
+}
+
+func minTimeToVisitAllPoints(points [][]int) (ans int) {
+    for i := 1; i < len(points); i++ {
+        ans += max(abs1266(points[i][0] - points[i - 1][0]), abs1266(points[i][1] - points[i - 1][1]))
+    }
+    return
+}
diff --git a/Codes/1266-minimum-time-visiting-all-points.java b/Codes/1266-minimum-time-visiting-all-points.java
@@ -0,0 +1,15 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-01-12 23:24:06
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-01-12 23:32:59
+ */
+class Solution {
+    public int minTimeToVisitAllPoints(int[][] points) {
+        int ans = 0;
+        for (int i = 1; i < points.length; i++) {
+            ans += Math.max(Math.abs(points[i][0] - points[i - 1][0]) , Math.abs(points[i][1] - points[i - 1][1]));
+        }
+        return ans;
+    }
+}
diff --git a/Codes/1266-minimum-time-visiting-all-points.py b/Codes/1266-minimum-time-visiting-all-points.py
@@ -0,0 +1,12 @@
+'''
+Author: LetMeFly
+Date: 2026-01-12 23:24:06
+LastEditors: LetMeFly.xyz
+LastEditTime: 2026-01-12 23:32:26
+'''
+from typing import List
+from itertools import pairwise
+
+class Solution:
+    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
+        return sum(max(abs(a[0] - b[0]), abs(a[1] - b[1])) for a, b in pairwise(points))
diff --git a/Codes/1266-minimum-time-visiting-all-points.rs b/Codes/1266-minimum-time-visiting-all-points.rs
@@ -0,0 +1,15 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-01-12 23:24:06
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-01-12 23:37:10
+ */
+impl Solution {
+    pub fn min_time_to_visit_all_points(points: Vec<Vec<i32>>) -> i32 {
+        let mut ans: i32 = 0;
+        for i in 1..points.len() {
+            ans += (points[i][0] - points[i - 1][0]).abs().max((points[i][1] - points[i - 1][1]).abs());
+        }
+        ans
+    }
+}
diff --git a/Codes/lib.rs b/Codes/lib.rs
@@ -5,7 +5,7 @@
  * @LastEditTime: 2026-01-06 13:17:16
  */
 pub struct Solution;
-include!("1458-max-dot-product-of-two-subsequences.rs");  // 这个fileName是会被脚本替换掉的
+include!("1266-minimum-time-visiting-all-points.rs");  // 这个fileName是会被脚本替换掉的
 
 #[derive(Debug, PartialEq, Eq)]
 pub struct TreeNode {
diff --git a/README.md b/README.md
@@ -533,6 +533,7 @@
 |1260.二维网格迁移|简单|<a href="https://leetcode.cn/problems/shift-2d-grid/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/07/20/LeetCode%201260.%E4%BA%8C%E7%BB%B4%E7%BD%91%E6%A0%BC%E8%BF%81%E7%A7%BB/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/125889225" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/shift-2d-grid/solution/by-tisfy-7fsm/" target="_blank">LeetCode题解</a>|
 |1261.在受污染的二叉树中查找元素|中等|<a href="https://leetcode.cn/problems/find-elements-in-a-contaminated-binary-tree/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/03/12/LeetCode%201261.%E5%9C%A8%E5%8F%97%E6%B1%A1%E6%9F%93%E7%9A%84%E4%BA%8C%E5%8F%89%E6%A0%91%E4%B8%AD%E6%9F%A5%E6%89%BE%E5%85%83%E7%B4%A0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/136642314" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-elements-in-a-contaminated-binary-tree/solutions/2681786/letmefly-1261zai-shou-wu-ran-de-er-cha-s-fcb0/" target="_blank">LeetCode题解</a>|
 |1262.可被三整除的最大和|中等|<a href="https://leetcode.cn/problems/greatest-sum-divisible-by-three/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/06/19/LeetCode%201262.%E5%8F%AF%E8%A2%AB%E4%B8%89%E6%95%B4%E9%99%A4%E7%9A%84%E6%9C%80%E5%A4%A7%E5%92%8C/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131290984" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/greatest-sum-divisible-by-three/solutions/2314201/letmefly-1262ke-bei-san-zheng-chu-de-zui-wyt8/" target="_blank">LeetCode题解</a>|
+|1266.访问所有点的最小时间|简单|<a href="https://leetcode.cn/problems/minimum-time-visiting-all-points/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/01/12/LeetCode%201266.%E8%AE%BF%E9%97%AE%E6%89%80%E6%9C%89%E7%82%B9%E7%9A%84%E6%9C%80%E5%B0%8F%E6%97%B6%E9%97%B4/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/156875167" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-time-visiting-all-points/solutions/3879062/letmefly-1266fang-wen-suo-you-dian-de-zu-lorr/" target="_blank">LeetCode题解</a>|
 |1267.统计参与通信的服务器|中等|<a href="https://leetcode.cn/problems/count-servers-that-communicate/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/08/24/LeetCode%201267.%E7%BB%9F%E8%AE%A1%E5%8F%82%E4%B8%8E%E9%80%9A%E4%BF%A1%E7%9A%84%E6%9C%8D%E5%8A%A1%E5%99%A8/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132466649" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-servers-that-communicate/solutions/2402129/letmefly-1267tong-ji-can-yu-tong-xin-de-5r3r8/" target="_blank">LeetCode题解</a>|
 |1276.不浪费原料的汉堡制作方案|中等|<a href="https://leetcode.cn/problems/number-of-burgers-with-no-waste-of-ingredients/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/12/25/LeetCode%201276.%E4%B8%8D%E6%B5%AA%E8%B4%B9%E5%8E%9F%E6%96%99%E7%9A%84%E6%B1%89%E5%A0%A1%E5%88%B6%E4%BD%9C%E6%96%B9%E6%A1%88/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135196793" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-burgers-with-no-waste-of-ingredients/solutions/2579080/letmefly-1276bu-lang-fei-yuan-liao-de-yi-b1t2/" target="_blank">LeetCode题解</a>|
 |1278.分割回文串III|困难|<a href="https://leetcode.cn/problems/palindrome-partitioning-iii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/03/LeetCode%201278.%E5%88%86%E5%89%B2%E5%9B%9E%E6%96%87%E4%B8%B2III/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/145986494" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/palindrome-partitioning-iii/solutions/3593573/letmefly-1278fen-ge-hui-wen-chuan-iiidon-c0va/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 1266.访问所有点的最小时间.md b/Solutions/LeetCode 1266.访问所有点的最小时间.md
@@ -0,0 +1,167 @@
+---
+title: 1266.访问所有点的最小时间：贪心（数学）+python一行版
+date: 2026-01-12 23:37:50
+tags: [题解, LeetCode, 简单, 几何, 数组, 数学, 贪心]
+categories: [题解, LeetCode]
+index_img: https://assets.leetcode.cn/aliyun-lc-upload/uploads/2019/11/24/1626_example_1.png
+---
+
+# 【LetMeFly】1266.访问所有点的最小时间：贪心（数学）+python一行版
+
+力扣题目链接：[https://leetcode.cn/problems/minimum-time-visiting-all-points/](https://leetcode.cn/problems/minimum-time-visiting-all-points/)
+
+<p>平面上有 <code>n</code> 个点，点的位置用整数坐标表示 <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> 。请你计算访问所有这些点需要的 <strong>最小时间</strong>（以秒为单位）。</p>
+
+<p>你需要按照下面的规则在平面上移动：</p>
+
+<ul>
+	<li>每一秒内，你可以：
+	<ul>
+		<li>沿水平方向移动一个单位长度，或者</li>
+		<li>沿竖直方向移动一个单位长度，或者</li>
+		<li>跨过对角线移动 <code>sqrt(2)</code> 个单位长度（可以看作在一秒内向水平和竖直方向各移动一个单位长度）。</li>
+	</ul>
+	</li>
+	<li>必须按照数组中出现的顺序来访问这些点。</li>
+	<li>在访问某个点时，可以经过该点后面出现的点，但经过的那些点不算作有效访问。</li>
+</ul>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.cn/aliyun-lc-upload/uploads/2019/11/24/1626_example_1.png" style="height: 428px; width: 500px;" /></p>
+
+<pre>
+<strong>输入：</strong>points = [[1,1],[3,4],[-1,0]]
+<strong>输出：</strong>7
+<strong>解释：</strong>一条最佳的访问路径是： <strong>[1,1]</strong> -> [2,2] -> [3,3] -> <strong>[3,4] </strong>-> [2,3] -> [1,2] -> [0,1] -> <strong>[-1,0]</strong>   
+从 [1,1] 到 [3,4] 需要 3 秒 
+从 [3,4] 到 [-1,0] 需要 4 秒
+一共需要 7 秒</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>points = [[3,2],[-2,2]]
+<strong>输出：</strong>5
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>points.length == n</code></li>
+	<li><code>1 <= n <= 100</code></li>
+	<li><code>points[i].length == 2</code></li>
+	<li><code>-1000 <= points[i][0], points[i][1] <= 1000</code></li>
+</ul>
+
+
+    
+## 解题方法：贪心
+
+斜着移动一次相当于横着移动一次加竖着移动一次，点的访问顺序是固定的，所以从a点到b点时：
+
+> 我们尽可能多地斜向移动，移动到一条水平线或竖直线时水平/竖直移动。
+>
+> 总移动次数：$max(水平diff, 竖直diff)$。相当于斜向移动时候把近的那个水平/竖直分量给一块走了。
+
++ 时间复杂度$O(len(pints))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2026-01-12 23:28:12
+ */
+class Solution {
+public:
+    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
+        int ans = 0;
+        for (int i = 1; i < points.size(); i++) {
+            ans += max(abs(points[i][0] - points[i - 1][0]), abs(points[i][1] - points[i - 1][1]));
+        }
+        return ans;
+    }
+};
+```
+
+#### Python
+
+```python
+'''
+LastEditTime: 2026-01-12 23:32:26
+'''
+from typing import List
+from itertools import pairwise
+
+class Solution:
+    def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int:
+        return sum(max(abs(a[0] - b[0]), abs(a[1] - b[1])) for a, b in pairwise(points))
+```
+
+#### Java
+
+```java
+/*
+ * @LastEditTime: 2026-01-12 23:32:59
+ */
+class Solution {
+    public int minTimeToVisitAllPoints(int[][] points) {
+        int ans = 0;
+        for (int i = 1; i < points.length; i++) {
+            ans += Math.max(Math.abs(points[i][0] - points[i - 1][0]) , Math.abs(points[i][1] - points[i - 1][1]));
+        }
+        return ans;
+    }
+}
+```
+
+#### Go
+
+```go
+/*
+ * @LastEditTime: 2026-01-12 23:35:23
+ */
+package main
+
+func abs1266(a int) int {
+    if a < 0 {
+        return -a
+    }
+    return a
+}
+
+func minTimeToVisitAllPoints(points [][]int) (ans int) {
+    for i := 1; i < len(points); i++ {
+        ans += max(abs1266(points[i][0] - points[i - 1][0]), abs1266(points[i][1] - points[i - 1][1]))
+    }
+    return
+}
+```
+
+#### Rust
+
+```rust
+/*
+ * @LastEditTime: 2026-01-12 23:37:10
+ */
+impl Solution {
+    pub fn min_time_to_visit_all_points(points: Vec<Vec<i32>>) -> i32 {
+        let mut ans: i32 = 0;
+        for i in 1..points.len() {
+            ans += (points[i][0] - points[i - 1][0]).abs().max((points[i][1] - points[i - 1][1]).abs());
+        }
+        ans
+    }
+}
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/156875167)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2026/01/12/LeetCode%201266.%E8%AE%BF%E9%97%AE%E6%89%80%E6%9C%89%E7%82%B9%E7%9A%84%E6%9C%80%E5%B0%8F%E6%97%B6%E9%97%B4/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/Other-English-LearningNotes-SomeWords.md b/Solutions/Other-English-LearningNotes-SomeWords.md
