567. 字符串的排列

发布时间 2023-03-31 14:52:54作者: Co3y

力扣题目链接

给你两个字符串 s1 和 s2 ,写一个函数来判断 s2 是否包含 s1 的排列。如果是,返回 true ;否则,返回 false 。

换句话说,s1 的排列之一是 s2 的 子串 。

示例 1:

输入:s1 = "ab" s2 = "eidbaooo"
输出:true
解释:s2 包含 s1 的排列之一 ("ba").

示例 2:

输入:s1= "ab" s2 = "eidboaoo"
输出:false

提示:

  • 1 <= s1.length, s2.length <= 104
  • s1 和 s2 仅包含小写字母
class Solution {
    public boolean checkInclusion(String s1, String s2) {
        HashMap<Character, Integer> window = new HashMap<>();
        HashMap<Character, Integer> need = new HashMap<>();
        for (char c : s1.toCharArray()) {
            need.put(c, need.getOrDefault(c, 0) + 1);
        }
        int left = 0, right = 0;
        int valid = 0;
        while (right < s2.length()) {
            char c = s2.charAt(right);
            right++;
            if (need.containsKey(c)) {
                window.put(c, window.getOrDefault(c, 0) + 1);
                //包装类需要使用equals判断是否相等
                if (window.get(c).equals(need.get(c)))
                    valid++;
            }

            //System.out.println("window: [" + left + "," + right + ")");

            if (right - left == s1.length()) {
                if (valid == need.size()) {
                    return true;
                }
                char d = s2.charAt(left);
                left++;
                if (need.containsKey(d)) {
                    //只有先判断相等valid才能减1
                    if (window.get(d).equals(need.get(d)))
                        valid--;
                    window.put(d, window.get(d) - 1);
                }
            }
        }
        return false;
    }
}