Sum of Beauty of All Substrings

给一个string, 求它所有的substring的最大字符count和最小字符count的差.

这个没什么可优化的, 简单的先处理下substring的count方法即可.

class Solution {
    public int beautySum(String s) {
        int[] count = new int[26];
        int res = 0;
        for(int i = 0; i < s.length(); i++) {
            count[s.charAt(i) - 'a']++; 
            for(int j = i + 1; j < s.length(); j++) {
                count[s.charAt(j) - 'a']++;
                res += find(count);
            }
            count = new int[26];
        }
        return res;
    }
    public int find(int[] count) {
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for(int c : count){
            if(c == 0)
                continue;
            min = Math.min(min, c);
            max = Math.max(max, c);
        } 
        return max - min;
    }
}