LeetCode--179场周赛题解

水题：

class Solution {
public:
    string generateTheString(int n) {
    string s;
    string a="a",b="b";
    if( n%2==0){
        
        for(int i =1; i <= n - 1; i++)
            s = s + a;
        s = s + b;
    }
    else{
        for(int i =1; i <= n ; i++)
            s = s + a;
    }
        return s;
    }
};

难题：1377. T 秒后青蛙的位置

思路：自底向上反推，从结果入手寻找答案

#include<string.h>
class Solution {
public:
    double frogPosition(int n, vector< vector<int> >& edges, int t, int target) {
        
        int a[105];
        int step = t;
        double ans = 1;
        memset(a, 0, sizeof(a) );
        if (edges.size() == 0 ){
            return 1;
        }
        else if ( target == 1) {
            return 0;
        }
        
         for (int i = 0; i < edges.size(); i++){
             if (edges[i][0]>edges[i][1]){
                 swap(edges[i][0],edges[i][1]);
             }
             a[edges[i][0]]++;
             
        }
        
        int flag = a[target];
        
        while(1){
            for (int i = 0; i < edges.size(); i++) {

                if ( edges[i][1] == target){
                      
                        target = edges[i][0];
                        ans = ans /a[edges[i][0]];
                        t--;
                         
                        if (target == 1) {
                            if ( t == 0)  return ans;
                            else if (t > 0 && flag == 0) return ans;
                            else return 0;
                        }
                    break;
                                 
                 }

                

            }
        }
    }
};