Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

问题：大数据的时候TLE,问题搜索的时候，走了很多重复的子路径

所以不能进行DFS(DFS本质上是暴力),而应该DP

DP[i,j] = 代表从A[0,0]到A[i,j]的path个数

DP[i,j] = DP[i-1,j] + DP[i,j-1] 接下来自底向上或者memorize的方式都OK

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > dp = grid;
        
        for(int i = 0; i < grid.size(); i++){
            for(int j = 0; j < grid[0].size();j++){
                if (i > 0 && j > 0){
                    dp[i][j] = min(dp[i][j -1],dp[i -1][j]) + grid[i][j];
                }else if (i > 0){
                    dp[i][j] = dp[i -1][j] + grid[i][j];
                }else if (j > 0){
                    dp[i][j] = dp[i][j -1] + grid[i][j];
                }else{
                    dp[i][j] = grid[i][j];
                }
            }
        }
        
        return dp[grid.size() -1][grid[0].size() -1];
    }
};

 

class Solution {
public:
    void dp1(int m, int n){
        vector<int> dp(n.size(),1);
        for(int i = 1; i < m; i++){
            for(int j = 0; j < n; j++){
                //dp[i][j] = dp[i-1][j] + dp[i][j-1];
                if (j) dp[j] = dp[j] + dp[j-1];
            }
        }
        return dp[dp.size() -1];
    }
    //继续优化到一个2个变量，问题是需要保存上一次的i-1的dp[j]的值，而用ans pre两个变量只能保存到当前的i的dp[j]的值
    //继续优化失败
    void dp2(int m, int n){
        int ans = 1, pre = 0;
        for(int i = 1; i < m; i++){
            for(int j = 0; j < n; j++){
                if (j){
                     ans = ans + pre;
                }else{
                    ans = 1;
                }
                pre = ans;
            }
        }
        return ans;
    }
    int uniquePaths(int m, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return dp1(m,n);
    }
};