Swordsman Offer Special Assault Edition ---- Day 6

class Solution {
    
    public String minWindow(String s, String t) {
    
        int n = s.length(),m = t.length();
        if(n<m) return "";
        int[] cnt = new int[150];
        int k = 0;
        for(char c:t.toCharArray()){
    
            if(++cnt[c]==1){
    
                k++;
            }
        }
        int l = 0,left = 0,right = 0;
        for(int i = 0;i<s.length();i++){
    
            char c = s.charAt(i);
            if(--cnt[c]==0) k--;
            if(k==0){
    
                while(cnt[s.charAt(l)]<0){
    
                    ++cnt[s.charAt(l++)];
                }
                if(left==right||right-left>i-l+1){
    
                    right = i+1;
                    left = l;
                }
            }
        }
        return s.substring(left,right);
    }
}

class Solution {
    
    public boolean isPalindrome(String s) {
    
        StringBuilder str = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
    
            char c = s.charAt(i);
            if (Character.isLetterOrDigit(c)) {
    
                str.append(Character.toLowerCase(c));
            }
        }
        return str.toString().equals(new StringBuffer(str).reverse().toString());
    }
}

class Solution {
    
    public boolean validPalindrome(String s) {
    
        for(int left = 0, right = s.length() - 1; left < right; left++, right--){
    
            // 如果不相等,则分两种情况：删除左边的元素,或者右边的元素,再判断各自是否回文,满足一种即可.
            if(s.charAt(left) != s.charAt(right)) 
                return isPalindrome(s, left+1, right) || isPalindrome(s, left, right - 1);
        }
        return true;
    }

    // 判断字符串 s 的 [left, right] 是否回文
    private boolean isPalindrome(String s, int left , int right){
    
        while (left < right){
    
            if (s.charAt(left++) != s.charAt(right--))
                return false;
        }
        return true;
    }
}

class Solution {
    
    public int countSubstrings(String s) {
    
        int n = s.length(), ans = 0;
        for (int i = 0; i < 2 * n - 1; ++i) {
    
            int l = i / 2, r = i / 2 + i % 2;
            while (l >= 0 && r < n && s.charAt(l) == s.charAt(r)) {
    
                --l;
                ++r;
                ++ans;
            }
        }
        return ans;
    }
}