19.2.11 [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.

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.
 1 class Solution {
 2 public:
 3     int caluniquePaths(vector<vector<int>>&grid,vector<vector<int>>&mark,int m, int n) {
 4         int down = -1, right = -1;
 5         if (mark[m][n] != -1)return mark[m][n];
 6         int orim = grid.size(), orin = grid[0].size();
 7         if (m > 1)
 8             down = caluniquePaths(grid,mark, m - 1, n);
 9         if (n > 1)
10             right = caluniquePaths(grid,mark, m, n - 1);
11         if (down == -1)down = right;
12         if (right == -1)right = down;
13         mark[m][n] = grid[orim-m][orin-n]+min(down,right);
14         return mark[m][n];
15     }
16     int minPathSum(vector<vector<int>>& grid) {
17         int m = grid.size(), n = grid[0].size();
18         vector<vector<int>>mark(m+1, vector<int>(n+1, -1));
19         mark[1][1] = grid[m-1][n-1];
20         return caluniquePaths(grid,mark, m, n);
21     }
22 };
View Code

用的改的上一题的题解，但偏慢，不如改成dp快，估计是调用比较慢

 1 class Solution {
 2 public:
 3     int minPathSum(vector<vector<int>>& grid) {
 4         int m = grid.size(), n = grid[0].size();
 5         vector<vector<int>>dp(m, vector<int>(n, INT_MAX));
 6         dp[0][0] = grid[0][0];
 7         for(int i=0;i<m;i++)
 8             for (int j = 0; j < n; j++) {
 9                 if (i == 0 && j == 0)continue;
10                 if (i == 0) {
11                     dp[i][j] = min(dp[i][j - 1] + grid[i][j], dp[i][j]);
12                 }
13                 else if (j == 0) {
14                     dp[i][j] = min(dp[i - 1][j] + grid[i][j], dp[i][j]);
15                 }
16                 else {
17                     dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j], dp[i][j]);
18                 }
19             }
20         return dp[m - 1][n - 1];
21     }
22 };
View Code
posted @ 2019-02-11 11:14  TobicYAL  阅读(177)  评论(0)  编辑  收藏  举报