leetcode || 64、Minimum Path Sum

problem:

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.

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 Array Dynamic Programming
题意:从矩阵的左上方到右下方寻找一条加权最小的路径

thinking:

(1)矩阵的路径问题(求路径总数。带障碍物的路径总数。加权最小、最大路径等),因为具有清晰且简易的状态转移公式,
统统能够用DP法解决,时间复杂度都为O(m*n)

(2) 用DP解决全局最优问题!

。该题是否具有局部最优解呢。状态转移公式:a[i][j] = min(a[i-1][j], a[i][j-1]) + grid[i][j];

规定了每一步的选择都是最优解。所以局部最优是成立的

code:

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) {
        vector<vector<int> >::const_iterator con_it=grid.begin();
        int m=grid.size();
        int n=(*con_it).size();
        vector<int> tmp(n,0);
        vector<vector<int> > a(m,tmp);
        if(m==0 || n==0)
            return 0;
        a[0][0]=grid[0][0];
        for(int i=1;i<m;i++)
           a[i][0]=a[i-1][0]+grid[i][0];
        for(int j=1;j<n;j++)
            a[0][j]=a[0][j-1]+grid[0][j];
        for(int i = 1; i < m; i++)
            for(int j = 1; j < n; j++)
                a[i][j] = min(a[i-1][j], a[i][j-1]) + grid[i][j];
        return a[m-1][n-1];
    }
};


原文地址:https://www.cnblogs.com/liguangsunls/p/6743798.html