滑动窗口问题的实现

滑动窗口的模板

int left = 0, right = 0;

while (right < s.size()) {`
    // 增大窗口
    window.add(s[right]);
    right++;
    
    while (window needs shrink) {
        // 缩小窗口
        window.remove(s[left]);
        left++;
    }
}

实例

• Leetcode 3 无重复字符的最长子串长度

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s.length() == 0 ){
            return 0 ;
        }
        HashMap<Character,Integer> map= new HashMap<Character,Integer>();
        int max =0;
        int left = 0 ;
        for(int i = 0 ;i< s.length(); ++i){
            if(map.containsKey(s.charAt(i))){
                left = Math.max(left,map.get(s.charAt(i))+1);
            }
            map.put(s.charAt(i),i);
            max=Math.max(max,i-left+1);
        }
        return max ;
    }
}

• LeetCode 76 最小覆盖子串

class Solution {
    public static String minWindow(String s, String t) {
        if (s == null || s == "" || t == null || t == "" || s.length() < t.length()) {
            return "";
        }
        //用来统计t中每个字符出现次数
        int[] needs = new int[128];
        //用来统计滑动窗口中每个字符出现次数
        int[] window = new int[128];

        for (int i = 0; i < t.length(); i++) {
            needs[t.charAt(i)]++;
        }

        int left = 0;
        int right = 0;

        String res = "";

        //目前有多少个字符
        int count = 0;

        //用来记录最短需要多少个字符。
        int minLength = s.length() + 1;

        while (right < s.length()) {
            char ch = s.charAt(right);
            window[ch]++;
            if (needs[ch] > 0 && needs[ch] >= window[ch]) {
                count++;
            }

            //移动到不满足条件为止
            while (count == t.length()) {
                ch = s.charAt(left);
                if (needs[ch] > 0 && needs[ch] >= window[ch]) {
                    count--;
                }
                if (right - left + 1 < minLength) {
                    minLength = right - left + 1;
                    res = s.substring(left, right + 1);

                }
                window[ch]--;
                left++;

            }
            right++;

        }
        return res;
    }

}