leetcode:76

题目描述

timumiaoshu

思路分析

采用滑动窗口,窗口有左右边界,先通过扩展右边界找出一个包含T中所有字符的子串,然后收缩左边界,直到不能再收缩。记录此时的子串。然后收缩左边界,继续扩展右边界,直到再找到满足要求的子串,和上次的进行比较,保存更小的子串。返回执行,直到右边界到达S串尾,且左边界不能再收缩。

代码

public static String minWindow(String s, String t) {
        Map<Character, Integer> map = new HashMap<>();
        int min = Integer.MAX_VALUE;
        int minStart = 0, minEnd = 0;
        int count = t.length();
        for (char c : t.toCharArray()) {
            map.put(c, map.containsKey(c) ? map.get(c) + 1 : 1);
        }
        int left = 0;
        for (int right = 0; right < s.length(); right++) {
            char val = s.charAt(right);
            if (map.containsKey(val)) {
                map.put(val, map.get(val) - 1);
                if (map.get(val) >= 0) {
                    count--;
                }
            }
            while (count == 0) {
                if (right - left < min) {
                    min = right - left;
                    minStart = left;
                    minEnd = right;
                }
                char temp = s.charAt(left);
                if (map.containsKey(temp)) {
                    map.put(temp, map.get(temp) + 1);
                    if (map.get(temp) > 0) count++;
                }
                left++;
            }
        }
        return min == Integer.MAX_VALUE ? "" : s.substring(minStart, minEnd + 1);

猜你喜欢

转载自blog.csdn.net/huangbx_tx/article/details/85225987
今日推荐