|
| 1 | +/* |
| 2 | + * @lc app=leetcode.cn id=76 lang=java |
| 3 | + * |
| 4 | + * [76] 最小覆盖子串 |
| 5 | + * |
| 6 | + * https://leetcode-cn.com/problems/minimum-window-substring/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Hard (37.77%) |
| 10 | + * Likes: 636 |
| 11 | + * Dislikes: 0 |
| 12 | + * Total Accepted: 61.7K |
| 13 | + * Total Submissions: 161.6K |
| 14 | + * Testcase Example: '"ADOBECODEBANC"\n"ABC"' |
| 15 | + * |
| 16 | + * 给你一个字符串 S、一个字符串 T,请在字符串 S 里面找出:包含 T 所有字符的最小子串。 |
| 17 | + * |
| 18 | + * 示例: |
| 19 | + * |
| 20 | + * 输入: S = "ADOBECODEBANC", T = "ABC" |
| 21 | + * 输出: "BANC" |
| 22 | + * |
| 23 | + * 说明: |
| 24 | + * |
| 25 | + * |
| 26 | + * 如果 S 中不存这样的子串,则返回空字符串 ""。 |
| 27 | + * 如果 S 中存在这样的子串,我们保证它是唯一的答案。 |
| 28 | + * |
| 29 | + * |
| 30 | + */ |
| 31 | + |
| 32 | +// @lc code=start |
| 33 | +class Solution { |
| 34 | + public String minWindow(String s, String t) { |
| 35 | + int[] need = new int[128]; |
| 36 | + int[] window = new int[128]; |
| 37 | + for (char ch : t.toCharArray()) { |
| 38 | + need[ch]++; |
| 39 | + } |
| 40 | + |
| 41 | + int left = 0, right = 0; |
| 42 | + int count = 0; |
| 43 | + int start = 0, len = s.length() + 1; |
| 44 | + |
| 45 | + while (right < s.length()) { |
| 46 | + char ch = s.charAt(right); |
| 47 | + right++; |
| 48 | + if (need[ch] > 0) { |
| 49 | + window[ch]++; |
| 50 | + if (window[ch] <= need[ch]) { |
| 51 | + count++; |
| 52 | + } |
| 53 | + } |
| 54 | + |
| 55 | + while (count == t.length()) { |
| 56 | + if (right - left < len) { |
| 57 | + start = left; |
| 58 | + len = right - left; |
| 59 | + } |
| 60 | + |
| 61 | + ch = s.charAt(left); |
| 62 | + left++; |
| 63 | + if (need[ch] > 0) { |
| 64 | + if (window[ch] <= need[ch]) { |
| 65 | + count--; |
| 66 | + } |
| 67 | + window[ch]--; |
| 68 | + } |
| 69 | + } |
| 70 | + } |
| 71 | + |
| 72 | + return (len > s.length()) ? "" : s.substring(start, start + len); |
| 73 | + } |
| 74 | +} |
| 75 | +// @lc code=end |
| 76 | + |
0 commit comments