leetcode 1573

简介

我们自己观察题目发现了什么这是一道数学题，哈哈哈。
个人的思路是分成两类去判断，
第一种：
全是0
使用

[ (n-1) * (n - 2) / 2 ]

第二种：
有1
然后观察10101
发现10101
其中0的个数两个之间
1个和1个

[ （前一堆1和中间一堆1之间的0的个数+1) * (后一堆1和中间一堆1之间的0的个数+1) = 结果 ]

Tips

注意防止溢出

AC code

class Solution1573 {
public:
    int numWays(string s) {
        // step 1. count the number of the 1
        long int numberOne = 0;
        for(long int i=0; i<s.length(); i++){
            if(s[i] == '1'){
                numberOne++;
            }
        }
        // step 2. check the numberOne can be subdivided 3?
        if(numberOne % 3 != 0){
            return 0;
        }
        if(numberOne == 0){
            double n = s.length() - 1;
            return (long int)((n / 2) * (n-1))  % (1000000000 + 7)  ;
        }
        long int onePiece = numberOne / 3;
        // step 3. get the numberZero between first piece and third piece
        long int numberZero1 = 0; long int numberZero2 = 0;
        long int index = 0;
        long int status = -1;
        for(long int i=0; i<s.length(); i++){
            if(index == onePiece){
                status = 0;
            }
            if(status == -1){
                if(s[i] == '1'){
                    index++;
                }
            }
            if(status == 0){
                if(s[i] == '1'){
                    status = 1;
                    break;
                }else if(s[i] == '0'){
                    numberZero1++;
                }
            }
        }
        string tmp;
        tmp.resize(s.length());
        long int j=0;
        for(long int i=s.length() -1; i>=0; i--,j++){
            tmp[j] = s[i]; 
        }
        index = 0;
        status = -1;
        for(long int i=0; i<tmp.length(); i++){
            if(index == onePiece){
                status = 0;
            }
            if(status == -1){
                if(tmp[i] == '1'){
                    index++;
                }
            }
            if(status == 0){
                if(tmp[i] == '1'){
                    status = 1;
                    break;
                }else if(tmp[i] == '0'){
                    numberZero2++;
                }
            }
        }
        // step 4. get the number
        return ((numberZero1+1) % (1000000000 + 7) * (numberZero2+1) % (1000000000 + 7))% (1000000000 + 7); 
    }
    long long factorial(long long n){
        long long num = 1;
        for(long long i=1; i <= n; i++){
            num = num * i % (1000000000 + 7);
        }
        return num;
    }
};

link

https://github.com/lishaohsuai/leetCode
https://github.com/haoel/leetcode