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50 lines
1.2 KiB
50 lines
1.2 KiB
package 第03期.mca_09;
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// 给你一个字符串 s 、一个字符串 t
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// 返回 s 中涵盖 t 所有字符的最小子串
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// 如果 s 中不存在涵盖 t 所有字符的子串，则返回空字符串 "" 。
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// 注意：对于 t 中重复字符，我们寻找的子字符串中该字符数量必须不少于 t 中该字符数量
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// 如果 s 中存在这样的子串，我们保证它是唯一的答案
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// 测试链接 : https://leetcode.cn/problems/minimum-window-substring/
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public class Code03_MinWindowLength {
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public static String minWindow(String s, String t) {
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if (s.length() < t.length()) {
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return "";
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}
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char[] str = s.toCharArray();
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char[] target = t.toCharArray();
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int[] map = new int[256];
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for (char cha : target) {
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map[cha]++;
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}
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int all = target.length;
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int L = 0;
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int R = 0;
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int minLen = Integer.MAX_VALUE;
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int ansl = -1;
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int ansr = -1;
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while (R != str.length) {
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map[str[R]]--;
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if (map[str[R]] >= 0) {
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all--;
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}
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if (all == 0) {
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while (map[str[L]] < 0) {
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map[str[L++]]++;
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}
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if (minLen > R - L + 1) {
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minLen = R - L + 1;
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ansl = L;
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ansr = R;
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}
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all++;
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map[str[L++]]++;
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}
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R++;
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}
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return minLen == Integer.MAX_VALUE ? "" : s.substring(ansl, ansr + 1);
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}
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}
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