update: 添加问题“3341.到达最后一个房间的最少时间I”的代码和题解(#923)

Signed-off-by: LetMeFly666 <[email protected]>

Author: LetMeFly666

Codes/3341-find-minimum-time-to-reach-last-room-i.cpp

@@ -0,0 +1,40 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-05-07 23:27:54
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-05-08 21:45:08
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+private:
+    static constexpr int directions[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
+public:
+    int minTimeToReach(vector<vector<int>>& moveTime) {
+        int n = moveTime.size(), m = moveTime[0].size();
+        vector<vector<int>> ans(n, vector<int>(m, 2000000000));
+        ans[0][0] = 0;
+        priority_queue<tuple<int, int, int>> pq;  // [<-t, x, y>, ...]
+        pq.push({0, 0, 0});
+        while (pq.size()) {
+            auto [t, x, y] = pq.top();
+            t = -t;
+            pq.pop();
+            for (int d = 0; d < 4; d++) {
+                int nx = x + directions[d][0];
+                int ny = y + directions[d][1];
+                if (nx < 0 || nx >= n || ny < 0 || ny >= m) {
+                    continue;
+                }
+                int nt = max(t, moveTime[nx][ny]) + 1;
+                if (nt < ans[nx][ny]) {
+                    ans[nx][ny] = nt;
+                    pq.push({-nt, nx, ny});
+                }
+            }
+        }
+        return ans[n - 1][m - 1];
+    }
+};

Codes/3341-find-minimum-time-to-reach-last-room-i.go

@@ -0,0 +1,57 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-05-07 23:27:54
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-05-08 22:19:42
+ */
+package main
+import "container/heap"
+
+var directions [][]int = [][]int{{0, 1}, {0, -1}, {1, 0}, {-1, 0}}
+
+func minTimeToReach(moveTime [][]int) int {
+    n, m := len(moveTime), len(moveTime[0])
+    ans := make([][]int, n)
+    for i := range ans {
+        ans[i] = make([]int, m)
+        for j := range ans[i] {
+            ans[i][j] = 2000000001
+        }
+    }
+    ans[0][0] = 0
+    pq := &pq3341{}
+    heap.Init(pq)
+    heap.Push(pq, node3341{0, 0, 0})
+    for len(*pq) > 0 {
+        node := heap.Pop(pq).(node3341)
+        t, x, y := node.t, node.x, node.y
+        if t > ans[x][y] {  // 注意不能是>=，因为入队时ans[x][y]会:=t
+            continue
+        }
+        for _, d := range directions {
+            nx := x + d[0]
+            ny := y + d[1]
+            if nx < 0 || nx >= n || ny < 0 || ny >= m {
+                continue
+            }
+            nt := max(t, moveTime[nx][ny]) + 1
+            if nt < ans[nx][ny] {
+                ans[nx][ny] = nt
+                heap.Push(pq, node3341{nt, nx, ny})
+            }
+        }
+    }
+    return ans[n - 1][m - 1]
+}
+
+type node3341 struct {
+    t, x, y int
+}
+
+type pq3341 []node3341
+
+func (pq *pq3341) Len() int           {return len(*pq)}
+func (pq *pq3341) Less(i, j int) bool {return (*pq)[i].t < (*pq)[j].t}
+func (pq *pq3341) Swap(i, j int)      {(*pq)[i], (*pq)[j] = (*pq)[j], (*pq)[i]}
+func (pq *pq3341) Push(node any)      {*pq = append(*pq, node.(node3341))}
+func (pq *pq3341) Pop() (ans any)     {*pq, ans = (*pq)[:len(*pq) - 1], (*pq)[len(*pq) - 1]; return ans}

Codes/3341-find-minimum-time-to-reach-last-room-i.java

@@ -0,0 +1,43 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-05-07 23:27:54
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-05-08 21:56:26
+ */
+import java.util.PriorityQueue;
+import java.util.Arrays;
+
+class Solution {
+    private final int[][] directions = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
+
+    public int minTimeToReach(int[][] moveTime) {
+        int n = moveTime.length, m = moveTime[0].length;
+        int[][] ans = new int[n][m];
+        for (int i = 0; i < n; i++) {
+            Arrays.fill(ans[i], 2000000001);
+        }
+        ans[0][0] = 0;
+        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
+        pq.offer(new int[]{0, 0, 0});
+        while (!pq.isEmpty()) {
+            int[] node = pq.poll();
+            int t = node[0], x = node[1], y = node[2];
+            if (t > ans[x][y]) {
+                continue;
+            }
+            for (int []d : directions) {
+                int nx = x + d[0];
+                int ny = y + d[1];
+                if (nx < 0 || nx >= n || ny < 0 || ny >= m) {
+                    continue;
+                }
+                int nt = Math.max(t, moveTime[nx][ny]) + 1;
+                if (nt < ans[nx][ny]) {
+                    ans[nx][ny] = nt;
+                    pq.offer(new int[]{nt, nx, ny});
+                }
+            }
+        }
+        return ans[n - 1][m - 1];
+    }
+}

Codes/3341-find-minimum-time-to-reach-last-room-i.py

@@ -0,0 +1,30 @@
+'''
+Author: LetMeFly
+Date: 2025-05-07 23:27:54
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-05-07 23:49:02
+'''
+from typing import List
+import heapq
+
+DIRECTIONS = [[0, 1], [0, -1], [1, 0], [-1, 0]]
+
+class Solution:
+    def minTimeToReach(self, moveTime: List[List[int]]) -> int:
+        n, m = len(moveTime), len(moveTime[0])
+        time = [[2000000000] * m for _ in range(n)]
+        time[0][0] = 0
+        pq = [(0, 0, 0)]
+        while pq:
+            t, x, y = heapq.heappop(pq)
+            if t > time[x][y]:
+                continue
+            for dx, dy in DIRECTIONS:
+                nx, ny = x + dx, y + dy
+                if not(0 <= nx < n and 0 <= ny < m):
+                    continue
+                nt = max(t, moveTime[nx][ny]) + 1
+                if nt < time[nx][ny]:
+                    time[nx][ny] = nt
+                    heapq.heappush(pq, (nt, nx, ny))
+        return time[n - 1][m - 1]

README.md

@@ -960,6 +960,7 @@
 |3305.元音辅音字符串计数I|中等|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/12/LeetCode%203305.%E5%85%83%E9%9F%B3%E8%BE%85%E9%9F%B3%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%AE%A1%E6%95%B0I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146197747" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-i/solutions/3607571/letmefly-3305yuan-yin-fu-yin-zi-fu-chuan-ptkg/" target="_blank">LeetCode题解</a>|
 |3306.元音辅音字符串计数II|中等|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/13/LeetCode%203306.%E5%85%83%E9%9F%B3%E8%BE%85%E9%9F%B3%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%AE%A1%E6%95%B0II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146228751" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-of-substrings-containing-every-vowel-and-k-consonants-ii/solutions/3609545/letmefly-3306yuan-yin-fu-yin-zi-fu-chuan-7y2q/" target="_blank">LeetCode题解</a>|
 |3340.检查平衡字符串|简单|<a href="https://leetcode.cn/problems/check-balanced-string/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/03/14/LeetCode%203340.%E6%A3%80%E6%9F%A5%E5%B9%B3%E8%A1%A1%E5%AD%97%E7%AC%A6%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/146249653" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/check-balanced-string/solutions/3610899/letmefly-3340jian-cha-ping-heng-zi-fu-ch-p8eo/" target="_blank">LeetCode题解</a>|
+|3341.到达最后一个房间的最少时间I|中等|<a href="https://leetcode.cn/problems/find-minimum-time-to-reach-last-room-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/05/07/LeetCode%203341.%E5%88%B0%E8%BE%BE%E6%9C%80%E5%90%8E%E4%B8%80%E4%B8%AA%E6%88%BF%E9%97%B4%E7%9A%84%E6%9C%80%E5%B0%91%E6%97%B6%E9%97%B4I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/CANNOT_PUBLISH" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-minimum-time-to-reach-last-room-i/solutions/3672076/letmefly-3341dao-da-zui-hou-yi-ge-fang-j-3m6o/" target="_blank">LeetCode题解</a>|
 |3375.使数组的值全部为K的最少操作次数|简单|<a href="https://leetcode.cn/problems/minimum-operations-to-make-array-values-equal-to-k/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/09/LeetCode%203375.%E4%BD%BF%E6%95%B0%E7%BB%84%E7%9A%84%E5%80%BC%E5%85%A8%E9%83%A8%E4%B8%BAK%E7%9A%84%E6%9C%80%E5%B0%91%E6%93%8D%E4%BD%9C%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147104288" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-operations-to-make-array-values-equal-to-k/solutions/3646138/letmefly-3375shi-shu-zu-de-zhi-quan-bu-w-i71e/" target="_blank">LeetCode题解</a>|
 |3392.统计符合条件长度为3的子数组数目|简单|<a href="https://leetcode.cn/problems/count-subarrays-of-length-three-with-a-condition/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/29/LeetCode%203392.%E7%BB%9F%E8%AE%A1%E7%AC%A6%E5%90%88%E6%9D%A1%E4%BB%B6%E9%95%BF%E5%BA%A6%E4%B8%BA3%E7%9A%84%E5%AD%90%E6%95%B0%E7%BB%84%E6%95%B0%E7%9B%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147603015" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-subarrays-of-length-three-with-a-condition/solutions/3665081/letmefly-3392tong-ji-fu-he-tiao-jian-cha-h935/" target="_blank">LeetCode题解</a>|
 |3396.使数组元素互不相同所需的最少操作次数|简单|<a href="https://leetcode.cn/problems/minimum-number-of-operations-to-make-elements-in-array-distinct/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/08/LeetCode%203396.%E4%BD%BF%E6%95%B0%E7%BB%84%E5%85%83%E7%B4%A0%E4%BA%92%E4%B8%8D%E7%9B%B8%E5%90%8C%E6%89%80%E9%9C%80%E7%9A%84%E6%9C%80%E5%B0%91%E6%93%8D%E4%BD%9C%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147080178" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-number-of-operations-to-make-elements-in-array-distinct/solutions/3644951/letmefly-3396shi-shu-zu-yuan-su-hu-bu-xi-glg4/" target="_blank">LeetCode题解</a>|