[leetcode 64] 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.

思路：dp，  dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]

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

在网上看到了大神的解法，将二维dp转为了一维。因为在计算dp[i][j]的时候我们要用的数据也只是dp[i-1][j] 和dp[i][j-1]。所以可以节省空间。

class Solution {public:    int minPathSum(vector<vector<int> > &grid) {        int res;        const int m = grid.size();        if (m < 1) {            return res;        }        const int n = grid[0].size();        int dp[n];        fill(dp, dp+n, INT_MAX);        dp[0] = 0;        for (int i = 0; i < m; i++) {            dp[0] += grid[i][0];            for (int j = 1; j < n; j++) {                //左边的dp[j]相当于二维中的d[i][j],右边的dp[j]相当于dp[i-1][j]，右边的dp[j-1]相当于dp[i][j-1];自己动手写2*n的就能看出来了                dp[j] = min(dp[j-1], dp[j]) + grid[i][j];            }        }        return dp[n-1];    }};