64. Minimum Path Sum

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题目

  • 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:

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

    解题思路

  • 本题比较简单，题意是寻找m x n格中左上角到右下角路径的最小距离，并且没有负权值
  • dp[i][j]:起点到点 [i,j]的最小距离
  • 因为没有负路径和寻路方向只能向下或者向右，因此对于每个点(除起点)，其dp[i][j]的表示为 dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j]

实现代码

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class Solution {
public:
    int minPathSum(vector<vector<int> >& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<int> > dp(m, vector<int>(n, 0));
        dp[0][0] = grid[0][0];
        for(int i = 0; i < m; ++i)
            for(int j = 0; j < n ; ++j) {
                if(i == 0 && j == 0)    continue;
                        if(i == 0)
                    dp[i][j] = dp[i][j-1] + grid[i][j];
                else if(j == 0)
                    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[m-1][n-1];
    }
};