LeetCode·64. Minimum Path Sum

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 1:

[[1,3,1], [1,5,1], [4,2,1]]

Given the above grid map, return 7. Because the path 1→3→1→1→1 minimizes the sum.

题目分析：

设一个dp的二维数组，然后分i，j循环来走

for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {     if (i == 0&&j==0) dp[0][0] = grid[0][0];                else 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];            }        }

代码：

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