int balancedStringSplit(char * s){ int count=0,num = 0; for (int i=0; i<=strlen(s); i++) { (s[i] == ‘R‘) ? count++:count--; if (!count) num++; } return num; }
【leetcode】分割平衡字符串
原文：https://www.cnblogs.com/ganxiang/p/13617737.html
