Minimum Path Sum
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Minimum Path Sum
- 问题描述:leetcode:64
- 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.
- 给定一个m*n的矩阵,从左上角找一个到右下角的最短路径。
思路:
- 对于一点dp[i][j]到达其最短路径为从其上方dp[i][j-1]或者从其左方到达dp[i-1][j];所以对于dp[i][j]:
dp[i][j] += min(dp[i-1][j] , dp[i][j-1]);
- 考虑边界情况:
对于左边界:只能从上方到达:
dp[i][0] += dp[i-1][0];
对于上边界:只能从左方到达:
dp[0][i] += dp[0][i-1];
代码:
int minPathSum(vector<vector<int>>& grid) { vector<vector<int> > &dp = grid; int row = dp.size(); int col = dp[0].size(); for (int i = 1; i<row; i++) dp[i][0] += dp[i-1][0]; for (int i = 1; i<col; i++) dp[0][i] += dp[0][i-1]; for (int i=1; i<row; i++){ for (int j=1; j<col; j++) { dp[i][j] += min(dp[i-1][j] , dp[i][j-1]); } } return dp[row-1][col-1];}
测试:
int main() { vector<vector<int> > b(3); int a[] = {2,4,3,7}; int a1[] = {5,3,2,1}; int a2[] = {4,8,6,2}; b[0].assign(a,a+4); b[1].assign(a1,a1+4); b[2].assign(a2,a2+4); printf("%d\n",minPathSum(b)); return 0;}
输出:14
语法注意点:
- 使用vector定义二维数组:
vector<vector<int> > b(3);
定义了一个二维数组b,有3行;列数未知。
注意:最后两个 > >中间有个空格。
vector定义的二维数组的行数:
int row = dp.size();
vector定义的二维数组的列数:
int col = dp[0].size();
vector定义的二维数组的赋值:
vector<vector<int> > b(3);int a[] = {2,4,3,7};int a1[] = {5,3,2,1};int a2[] = {4,8,6,2};b[0].assign(a,a+4);b[1].assign(a1,a1+4);b[2].assign(a2,a2+4);
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