动态规划Day01

1. 动态规划：63.不同路径II

   class Solution {
   public:
       int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
           int m = obstacleGrid.size();
           int n = obstacleGrid[0].size();
           if(obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n -1] == 1) {
               return 0;
           }
           vector<vector<int>>dp(m,vector<int>(n,0));
   
           for(int i = 0; i < m && obstacleGrid[i][0] == 0; i ++ ) dp[i][0] = 1;
           for(int j = 0; j < n && obstacleGrid[0][j] == 0; j ++ ) dp[0][j] = 1;
   
           for(int i = 1 ; i < m ; i ++ ) {
               for(int j = 1; j < n; j ++ ) {
                   if(obstacleGrid[i][j] == 1) continue;
                   dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
               }
           }
           return dp[m -1][n -1];
       }
   };

   五步走：1确定dpi的意义2找递推公式3确定初始值4确定遍历方向5代值验算

2. 动态规划：343.整数拆分

   class Solution {
   public:
       int integerBreak(int n) {
           vector<int>dp(n+1);
           dp[2] = 1;
           for(int i = 3; i <= n; i ++ ) {
               for(int j = 1; j <= i/2; j ++ ) {
                   dp[i] = max(dp[i],max(j * (i - j),j * dp[i - j]));
               }
           }
           return dp[n];
       }
   };

   3，如何确定大小端：

   #include <iostream>
   #include <cstring>
   
   using namespace std;
   
   bool test() {
       union {
           uint32_t i;
           uint8_t c[4];
       }u;
       u.i = 1;
       return u.c[0] == 1;
   }
   
   int main() {
       if(test()) {
           cout<<"small"<<endl;
       } else {
           cout <<"big" <<endl;
       }
       return 0;
   }