标签：path rom span time any sum nim from otto
题目：
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.
题解：
Solution 1 ()
class Solution { public: int minPathSum(vector<vector<int>>& grid) { int m = (int) grid.size(), n = (int) grid[0].size(); vector<long> dp(n,INT_MAX); dp[0] = grid[0][0]; for(int i=0; i<m; ++i) { for(int j=0; j<n; ++j) { if(j > 0) dp[j] = min(dp[j] + grid[i][j], dp[j-1] + grid[i][j]); else if(i>0) dp[j] = dp[j] + grid[i][j]; } } return dp[n-1]; } };
边界的第二种处理方法：因为Solution 1 中dp初始化为最大值，故需要考虑溢出情况，所以用long整型。这个就初始化为int整型。
Solution 2 ()
class Solution { public: int minPathSum(vector<vector<int>>& nums) { int m = (int) nums.size(); int n = (int) nums[0].size(); vector<int> v (n,0); for (int i=0; i<m; ++i) { for (int j=0; j<n; ++j) { if (i>0) v[j] = nums[i][j] + ((j>0) ? min(v[j], v[j-1]) : v[j]); else v[j] = nums[i][j] + ((j>0) ? v[j-1] : 0); } } return v[n-1]; } };
解法没变，就是边界的处理上不一样，这个是先初始化边界了。
Solution 3 ()
class Solution { public: int minPathSum(vector<vector<int>>& grid) { int dp[grid.size()][grid[0].size()]; dp[0][0] = grid[0][0]; // init first row for(int i = 1; i < grid[0].size(); i ++){ dp[0][i] = dp[0][i-1] + grid[0][i]; } // init first col for(int i = 1; i < grid.size(); i ++){ dp[i][0] = dp[i-1][0] + grid[i][0]; } for(int i = 1; i < grid.size(); i ++){ for(int j = 1; j < grid[0].size(); j++){ dp[i][j] = dp[i - 1][j] < dp[i][j-1]? dp[i - 1][j] + grid[i][j] : dp[i][j-1] + grid[i][j]; } } return dp[grid.size() - 1][grid[0].size() -1]; } };
【LeetCode】064. Minimum Path Sum
标签：path rom span time any sum nim from otto
原文地址：http://www.cnblogs.com/Atanisi/p/6817097.html
