From 29429f1b9864f947d76f096f2f4bc49899c80cb4 Mon Sep 17 00:00:00 2001 From: AC_Oier Date: Tue, 18 Jan 2022 09:13:21 +0800 Subject: [PATCH] =?UTF-8?q?=E2=9C=A8feat:=20Add=20539?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- "Index/\346\216\222\345\272\217.md" | 1 + "Index/\346\250\241\346\213\237.md" | 1 + ...10\344\270\255\347\255\211\357\274\211.md" | 190 ++++++++++++++++++ 3 files changed, 192 insertions(+) create mode 100644 "LeetCode/531-540/539. \346\234\200\345\260\217\346\227\266\351\227\264\345\267\256\357\274\210\344\270\255\347\255\211\357\274\211.md" diff --git "a/Index/\346\216\222\345\272\217.md" "b/Index/\346\216\222\345\272\217.md" index f23c49e8..5f873a01 100644 --- "a/Index/\346\216\222\345\272\217.md" +++ "b/Index/\346\216\222\345\272\217.md" @@ -8,6 +8,7 @@ | [475. 供暖器](https://leetcode-cn.com/problems/heaters/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/heaters/solution/gong-shui-san-xie-er-fen-shuang-zhi-zhen-mys4/) | 中等 | 🤩🤩🤩🤩 | | [506. 相对名次](https://leetcode-cn.com/problems/relative-ranks/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/relative-ranks/solution/gong-shui-san-xie-jian-dan-pai-xu-mo-ni-cmuzj/) | 简单 | 🤩🤩🤩🤩 | | [524. 通过删除字母匹配到字典里最长单词](https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/longest-word-in-dictionary-through-deleting/solution/gong-shui-san-xie-xiang-jie-pai-xu-shuan-qi20/) | 中等 | 🤩🤩🤩🤩 | +| [539. 最小时间差](https://leetcode-cn.com/problems/minimum-time-difference/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/minimum-time-difference/solution/gong-shui-san-xie-jian-dan-mo-ni-ti-by-a-eygg/) | 中等 | 🤩🤩🤩🤩 | | [581. 最短无序连续子数组](https://leetcode-cn.com/problems/shortest-unsorted-continuous-subarray/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/shortest-unsorted-continuous-subarray/solution/gong-shui-san-xie-yi-ti-shuang-jie-shuan-e1le/) | 中等 | 🤩🤩🤩🤩 | | [611. 有效三角形的个数](https://leetcode-cn.com/problems/valid-triangle-number/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/valid-triangle-number/solution/gong-shui-san-xie-yi-ti-san-jie-jian-dan-y1we/) | 中等 | 🤩🤩🤩🤩 | | [645. 错误的集合](https://leetcode-cn.com/problems/set-mismatch/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/set-mismatch/solution/gong-shui-san-xie-yi-ti-san-jie-ji-shu-s-vnr9/) | 简单 | 🤩🤩🤩 | diff --git "a/Index/\346\250\241\346\213\237.md" "b/Index/\346\250\241\346\213\237.md" index a75619c3..1c8cad3d 100644 --- "a/Index/\346\250\241\346\213\237.md" +++ "b/Index/\346\250\241\346\213\237.md" @@ -57,6 +57,7 @@ | [507. 完美数](https://leetcode-cn.com/problems/perfect-number/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/perfect-number/solution/gong-shui-san-xie-jian-dan-mo-ni-tong-ji-e6jk/) | 简单 | 🤩🤩🤩 | | [520. 检测大写字母](https://leetcode-cn.com/problems/detect-capital/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/detect-capital/solution/gong-shui-san-xie-jian-dan-zi-fu-chuan-m-rpor/) | 简单 | 🤩🤩🤩🤩 | | [528. 按权重随机选择](https://leetcode-cn.com/problems/random-pick-with-weight/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/random-pick-with-weight/solution/gong-shui-san-xie-yi-ti-shuang-jie-qian-8bx50/) | 中等 | 🤩🤩🤩🤩 | +| [539. 最小时间差](https://leetcode-cn.com/problems/minimum-time-difference/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/minimum-time-difference/solution/gong-shui-san-xie-jian-dan-mo-ni-ti-by-a-eygg/) | 中等 | 🤩🤩🤩🤩 | | [541. 反转字符串 II](https://leetcode-cn.com/problems/reverse-string-ii/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/reverse-string-ii/solution/gong-shui-san-xie-jian-dan-zi-fu-chuan-m-p88f/) | 简单 | 🤩🤩🤩🤩🤩 | | [551. 学生出勤记录 I](https://leetcode-cn.com/problems/student-attendance-record-i/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/student-attendance-record-i/solution/gong-shui-san-xie-jian-dan-mo-ni-ti-by-a-hui7/) | 简单 | 🤩🤩🤩 | | [566. 重塑矩阵](https://leetcode-cn.com/problems/reshape-the-matrix/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/reshape-the-matrix/solution/jian-dan-ti-zhong-quan-chu-ji-ke-yi-kan-79gv5/) | 简单 | 🤩🤩🤩 | diff --git "a/LeetCode/531-540/539. \346\234\200\345\260\217\346\227\266\351\227\264\345\267\256\357\274\210\344\270\255\347\255\211\357\274\211.md" "b/LeetCode/531-540/539. \346\234\200\345\260\217\346\227\266\351\227\264\345\267\256\357\274\210\344\270\255\347\255\211\357\274\211.md" new file mode 100644 index 00000000..c1a5574a --- /dev/null +++ "b/LeetCode/531-540/539. \346\234\200\345\260\217\346\227\266\351\227\264\345\267\256\357\274\210\344\270\255\347\255\211\357\274\211.md" @@ -0,0 +1,190 @@ +### 题目描述 + +这是 LeetCode 上的 **[539. 最小时间差](https://leetcode-cn.com/problems/minimum-time-difference/solution/gong-shui-san-xie-jian-dan-mo-ni-ti-by-a-eygg/)** ，难度为 **中等**。 + +Tag : 「模拟」、「排序」、「哈希表」 + + + +给定一个 $24$ 小时制（小时:分钟 `"HH:MM"`）的时间列表，找出列表中任意两个时间的最小时间差并以分钟数表示。 + +示例 1： +``` +输入：timePoints = ["23:59","00:00"] + +输出：1 +``` +示例 2： +``` +输入：timePoints = ["00:00","23:59","00:00"] + +输出：0 +``` + +提示： +* $2 <= timePoints <= 2 * 10^4$ +* `timePoints[i]` 格式为 `"HH:MM"` + +--- + +### 模拟（排序） + +根据题意，我们需要找出「时钟盘」中的夹角最小的两个时间点，其中包括了分布在 `00:00` 左右两侧（横跨了一天）的时间点。 + +因此，一种简单的方式是对于每个 $timePoints[i]$，我们不仅记录「当天该时间点」对应的的偏移量，还记录「下一天该时间点」对应的偏移量。 + +处理所有的 $timePoints[i]$ 后，对偏移量进行排序，遍历找到所有相邻元素之间的最小值。 + +**代码（感谢 [@Benhao](/u/himymben/) 和 [@🍭可乐可乐吗](/u/littletime_cc/) 同学提供的其他语言版本）：** +```Java +class Solution { + public int findMinDifference(List timePoints) { + int n = timePoints.size() * 2; + int[] nums = new int[n]; + for (int i = 0, idx = 0; i < n / 2; i++, idx += 2) { + String[] ss = timePoints.get(i).split(":"); + int h = Integer.parseInt(ss[0]), m = Integer.parseInt(ss[1]); + nums[idx] = h * 60 + m; + nums[idx + 1] = nums[idx] + 1440; + } + Arrays.sort(nums); + int ans = nums[1] - nums[0]; + for (int i = 0; i < n - 1; i++) ans = Math.min(ans, nums[i + 1] - nums[i]); + return ans; + } +} +``` +- +```Python3 +class Solution: + def findMinDifference(self, timePoints: List[str]) -> int: + n = len(timePoints) * 2 + nums, idx = [0] * n, 0 + for time in timePoints: + h, m = int(time[:2]), int(time[-2:]) + nums[idx] = h * 60 + m + nums[idx + 1] = nums[idx] + 1440 + idx += 2 + nums.sort() + return min(nums[i + 1] - nums[i] for i in range(n - 1)) +``` +- +```Go +func findMinDifference(timePoints []string) int { + n := len(timePoints) * 2 + nums := make([]int, n) + for i, idx := 0, 0; i < n / 2; i++ { + ss := strings.Split(timePoints[i], ":") + h, _ := strconv.Atoi(ss[0]) + m, _ := strconv.Atoi(ss[1]) + nums[idx] = h * 60 + m + nums[idx + 1] = nums[idx] + 1440 + idx += 2 + } + sort.Ints(nums) + ans := nums[1] - nums[0]; + for i := 1; i < n - 1; i++ { + if v := nums[i + 1] - nums[i]; v < ans { + ans = v + } + } + return ans +} +``` +- +```C++ +class Solution { +public: + int findMinDifference(vector& timePoints) { + int n = timePoints.size() * 2; + vector nums(n); + for(int i = 0, idx = 0; i < n / 2; i++, idx += 2){ + string ss = timePoints[i]; + auto pos = ss.find(':'); + int h = stoi(ss.substr(0, pos)), m = stoi(ss.substr(pos + 1)); + nums[idx] = h * 60 + m; + nums[idx + 1] = nums[idx] + 1440; + } + sort(nums.begin(), nums.end()); + int ans = nums[1] - nums[0]; + for(int i = 0; i < n - 1; i++){ + ans = min(ans, nums[i + 1] - nums[i]); + } + return ans; + } +}; +``` +- +```C +int min(int a,int b){ return a > b ? b : a; } +int cmp(const void* a , const void* b){ return (*(int*)a) - (*(int*)b); } + +int findMinDifference(char ** timePoints, int timePointsSize){ + int n = timePointsSize * 2; + int* nums = (int*) malloc(sizeof(int) * n); + for(int i = 0, idx = 0; i < n / 2; i++, idx += 2){ + timePoints[i][2] = '\0'; + int h = atoi(timePoints[i]), m = atoi(timePoints[i] + 3); + nums[idx] = h * 60 + m; + nums[idx + 1] = nums[idx] + 1440; + } + qsort(nums, n, sizeof(nums[0]), cmp); + int ans = nums[1] - nums[0]; + for(int i = 0; i < n - 1; i++){ + ans = min(ans, nums[i + 1] - nums[i]); + } + free(nums); + nums = NULL; + return ans; +} +``` +* 时间复杂度：$O(n\log{n})$ +* 空间复杂度：$O(n)$ + +--- + +### 模拟（哈希表计数） + +利用当天最多只有 $60 * 24 = 1440$ 个不同的时间点（跨天的话则是双倍），我们可以使用数组充当哈希表进行计数，同时根据「抽屉原理」，若 $timePoints$ 数量大于 $1440$，必然有两个相同时间点，用作剪枝。 + +然后找到间隔最小两个时间点，这种利用「桶排序」的思路避免了对 $timePoints$ 所对应的偏移量进行排序，而 $O(C)$ 的复杂度使得所能处理的数据范围没有上限。 + +代码： +```Java +class Solution { + public int findMinDifference(List timePoints) { + int n = timePoints.size(); + if (n >= 1440) return 0; + int[] cnts = new int[1440 * 2 + 10]; + for (String s : timePoints) { + String[] ss = s.split(":"); + int h = Integer.parseInt(ss[0]), m = Integer.parseInt(ss[1]); + cnts[h * 60 + m]++; + cnts[h * 60 + m + 1440]++; + } + int ans = 1440, last = -1; + for (int i = 0; i <= 1440 * 2 && ans != 0; i++) { + if (cnts[i] == 0) continue; + if (cnts[i] > 1) ans = 0; + else if (last != -1) ans = Math.min(ans, i - last); + last = i; + } + return ans; + } +} +``` +* 时间复杂度：$O(C)$ +* 空间复杂度：$O(C)$ + +--- + +### 最后 + +这是我们「刷穿 LeetCode」系列文章的第 `No.539` 篇，系列开始于 2021/01/01，截止于起始日 LeetCode 上共有 1916 道题目，部分是有锁题，我们将先把所有不带锁的题目刷完。 + +在这个系列文章里面，除了讲解解题思路以外，还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。 + +为了方便各位同学能够电脑上进行调试和提交代码，我建立了相关的仓库：https://github.com/SharingSource/LogicStack-LeetCode 。 + +在仓库地址里，你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解。 +