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  • 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];
        }
    };


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