【leetcode】分割平衡字符串

 

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;
}

 

posted @ 2020-09-05 12:07  温暖了寂寞  阅读(88)  评论(0编辑  收藏  举报