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.
典型的动态规划问题。
设dp[i][j]表示从左上角到grid[i][j]的最小路径和。那么dp[i][j] = grid[i][j] + min( dp[i-1][j], dp[i][j-1] );
下面的代码中，为了处理计算第一行和第一列的边界条件，我们令dp[i][j]表示从左上角到grid[i-1][j-1]的最小路径和，最后dp[m][n]是我们所求的结果
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class
Solution {public: int
minPathSum(vector<vector<int> > &grid) { int
row = grid.size(),col; if(row == 0)return
0; else
col = grid[0].size(); vector<vector<int> >dp(row+1, vector<int>(col+1, INT_MAX)); dp[0][1] = 0; for(int
i = 1; i <= row; i++) for(int
j = 1; j <= col; j++) dp[i][j] = grid[i-1][j-1] + min(dp[i][j-1], dp[i-1][j]); return
dp[row][col]; }}; |
注意到上面的代码中dp[i][j] 只和上一行的dp[i-1][j]和上一列的dp[i][j-1]有关，因此可以优化空间为O（n）（准确来讲空间复杂度可以是O（min（row,col
））） 本文地址
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class
Solution {public: int
minPathSum(vector<vector<int> > &grid) { int
row = grid.size(),col; if(row == 0)return
0; else
col = grid[0].size(); vector<int
>dp(col+1, INT_MAX); dp[1] = 0; for(int
i = 1; i <= row; i++) for(int
j = 1; j <= col; j++) dp[j] = grid[i-1][j-1] + min(dp[j], dp[j-1]); return
dp[col]; }}; |
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LeetCode:Minimum Path Sum,布布扣,bubuko.com
原文地址：http://www.cnblogs.com/TenosDoIt/p/3774804.html
