【LeetCode】Minimum Path Sum

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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.

 

解題思路：

典型的動態規劃。開闢m*n的矩陣minV，minV[i][j]存放從首元素(grid[0][0])到當前元素(grid[i][j])的最短路徑長度。

對於每個元素來說，路徑是從上或者從左邊來的。

也就是說minV[i][j] = min(minV[i-1][j]+minV[i][j-1]) + grid[i][j]。

別忘了初始化第一行第一列。

 

class Solution {public:    int minPathSum(vector<vector<int> > &grid)     {        int m = grid.size();        int n = grid[0].size();        //DP，存放從首元素到該元素的最短路徑        vector<vector<int> > minV;        minV.resize(m);        for(vector<vector<int> >::size_type st = 0; st < m; st ++)            minV[st].resize(n);        minV[0][0] = grid[0][0];        //第一列初始化        for(vector<vector<int> >::size_type st = 1; st < m; st ++)            minV[st][0] = minV[st-1][0] + grid[st][0];        //第一行初始化        for(vector<int>::size_type st = 1; st < n; st ++)            minV[0][st] = minV[0][st-1] + grid[0][st];        for(vector<vector<int> >::size_type st1 = 1; st1 < m; st1 ++)        {            for(vector<int>::size_type st2 = 1; st2 < n; st2 ++)            {                minV[st1][st2] = min(minV[st1-1][st2],minV[st1][st2-1])+grid[st1][st2];            }        }        return minV[m-1][n-1];    }};

【LeetCode】Minimum Path Sum