题解 | #扑克牌顺子#

#include <locale>
class Solution {
public:
    bool IsContinuous(vector<int>& numbers) {
        sort(numbers.begin(), numbers.end());
        int zero_num = 0;
        for(int i = 0; i < 5; i++){//首先计算数组中零的个数，以及判断是否有非零对子
            if(numbers[i] == 0) zero_num++;
           
            if(i != 0&& numbers[i] != 0 && numbers[i] == numbers[i -1]) return false;
        }
        
        for(int i = zero_num; i < 5; i++){
            if(i != zero_num && numbers[i] != numbers[i - 1] + 1){//如果两者不相邻
                if(zero_num == 0) return false; //但是此时零的数量已经用完，则返回false
                zero_num = zero_num - (numbers[i] - numbers[i - 1] - 1); //相当于少了这么对零的资源
                if(zero_num < 0) return false;
            }
        }
       
        return true;

    }
};