1. 题目描述
64. Minimum Path Sum
Medium
98225
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:
Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Because the path 1→3→1→1→1 minimizes the sum.
2.解题思路
dp数组记录到达(i,j)最短路径。因为题目规定只能向右走或者向下走,dp[ i ][ j ] = min ( dp[ i-1][ j ], dp[ i ][ j-1] ) + (i, j)本身的数值。要注意第一行的元素只能由其左边元素向右走到达,第一列元素只能由其上面元素向下走到达。
3.实现代码
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int row = grid.size(), col = grid[0].size();
int dp[row][col];
memset(dp, 0, sizeof(dp));
dp[0][0] = grid[0][0];
for (int i=1; i<col; i++) {//根据题意只能向下或向右走,所以第一行只能从(0,0)向右走到达
dp[0][i] = dp[0][i-1]+grid[0][i];
}
for (int i=1; i<row; i++) {
for (int j=0; j<col; j++) {
if (j==0)//第一列的元素只能由上一行第一列向下达到
dp[i][j] = dp[i-1][j] + grid[i][j];
else //非第一列元素正常比较哪一条路径更短即可
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
}
}
return dp[row-1][col-1];
}
};