64. Minimum Path Sum - 码农教程

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.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

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遍历的练手题,说是dp,其实跟暴力运算差不多了, 只是把暴力运算的中间结果存起来 ,空间换时间;

提交的时候提示那个faster than xx% 其实不准, 这题本来就比较耗时间, 纯遍历, O(m*n), 但实际程序运行的时候还要考虑cpu调度,内存分配等问题,与理论值相去甚远.

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        if(grid.empty()||grid[0].empty()) return 0;
        int h=grid.size(),w=grid[0].size();
        vector<vector<int>> dp(h,vector<int>(w));
        for(int i=0;i<h;++i)
            for(int j=0;j<w;++j)
            {
                if(0==i&&0==j) dp[i][j]=grid[i][j];
                else if (0==i&&j) dp[i][j]=dp[i][j-1]+grid[i][j];
                else if(0==j&&i) dp[i][j]+=dp[i-1][j]+grid[i][j];
                else dp[i][j]+=min(dp[i-1][j],dp[i][j-1])+grid[i][j];
            }
        return dp[h-1][w-1];
    }
};

原文地址：https://www.cnblogs.com/lychnis/p/11743120.html