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  • Leetcode 64. Minimum Path Sum

    64. Minimum Path Sum

    • Total Accepted: 76284
    • Total Submissions:214404
    • Difficulty: Medium

    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:优化的部分可以参考Leetcode 62. Unique Paths

    代码:
    方法一:未优化。

     1 class Solution {
     2 public:
     3     int minPathSum(vector<vector<int>>& grid) {
     4         int m=grid.size();
     5         int n=grid[0].size();
     6         int i,j;
     7         for(i=1;i<m;i++){
     8             grid[i][0]+=grid[i-1][0];
     9         }
    10         for(j=1;j<n;j++){
    11             grid[0][j]+=grid[0][j-1];
    12         }
    13         for(i=1;i<m;i++){
    14             for(j=1;j<n;j++){
    15                 grid[i][j]+=grid[i-1][j]<grid[i][j-1]?grid[i-1][j]:grid[i][j-1];
    16             }
    17         }
    18         return grid[m-1][n-1];
    19     }
    20 };

    方法二:已优化。

     1 class Solution {
     2 public:
     3     int minPathSum(vector<vector<int>>& grid) {
     4         int m=grid.size();
     5         int n=grid[0].size();
     6         int i,j;
     7         vector<int> cur(n,0);
     8         cur[0]=grid[0][0];
     9         for(j=1;j<n;j++){
    10             cur[j]=grid[0][j]+cur[j-1];
    11         }
    12         for(i=1;i<m;i++){
    13             cur[0]+=grid[i][0];
    14             for(j=1;j<n;j++){
    15                 cur[j]=grid[i][j]+(cur[j-1]>cur[j]?cur[j]:cur[j-1]);
    16             }
    17         }
    18         return cur[n-1];
    19     }
    20 };
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  • 原文地址:https://www.cnblogs.com/Deribs4/p/5654669.html
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