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如题：给定两个字符串 s 和 t 。返回 s 中包含 t 的所有字符的最短子字符串。如果 s 中不存在符合条件的子字符串，则返回空字符串 “” 。如果 s 中存在多个符合条件的子字符串，返回任意一个。

滑动窗口方案：

var minWindow = function (s, t) {if (!s || !s.length || !t || !t.length) return "";const len1 = t.length,len2 = s.length;const rObj = {};for (const c of t) {rObj[c] = (rObj[c] || 0) + 1;}let required = len1;let minStr = "";let minLen = Number.MAX_VALUE;let left = 0;for (let i = 0; i < len2; i++) {const cur = s[i];if (rObj[cur] !== void 0) {if (rObj[cur]-- > 0) required--;}while (left < i && (rObj[s[left]] === void 0 || rObj[s[left]] < 0)) {if (rObj[s[left]] < 0) rObj[s[left]]++;left++;}const len = i - left + 1;if (required == 0) {if (len < minLen) {minStr = s.slice(left, i + 1);minLen = len;}const leftChar = s[left++];if (rObj[leftChar] != void 0) {rObj[leftChar]++;required++;}}}return minStr;
};

优化：
使用更直观的变量名
优化滑动窗口收缩逻辑
简化条件判断
修复边界情况处理
使用现代 JavaScript 语法

var minWindow = function (s, t) {if (!s.length || !t.length) return "";const required = new Map(); // 需要匹配的字符计数for (const c of t) {required.set(c, (required.get(c) || 0) + 1);}let needCount = t.length; // 需要匹配的总字符数let minStr = ""; // 最小窗口字符串let minLen = Infinity; // 最小窗口长度let left = 0; // 滑动窗口左边界for (let right = 0; right < s.length; right++) {const char = s[right];// 处理当前字符if (required.has(char)) {const count = required.get(char);if (count > 0) needCount--; // 当字符仍有需求时才减少needCountrequired.set(char, count - 1);}// 当窗口满足条件时，尝试收缩左边界while (needCount === 0) {const currentLen = right - left + 1;// 更新最小窗口if (currentLen < minLen) {minLen = currentLen;minStr = s.slice(left, right + 1);}// 移动左边界并恢复状态const leftChar = s[left];if (required.has(leftChar)) {required.set(leftChar, required.get(leftChar) + 1);if (required.get(leftChar) > 0) needCount++;}left++;console.log('left,right==>', left,right);}}return minLen === Infinity ? "" : minStr;
};