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.
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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; } } |