Problem:
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.
Summary:
想要从m*n的整型数矩阵左上角走到右下角，每次只可以向右或向下移动一步，求路径上的整型数最小情况下的数字之和。
Solution:
1. 暴力法：枚举从左上角到右下角的所有路径，分别计算出路径和并比较。但效率过低且明显不可行，故不考虑。
2. 动态规划：由于只可以向右或向下移动，若已知(i, j)位置上一格(i - 1, j)以及左一格(i, j - 1)的路径数字之和，即可确定该位置的路径数字和：
dp[i][j] = min(dp[i - 1][j], dp[i][j - 1])
初始化：dp[0][0] = grid[0][0]
dp[0][j] = dp[0][j - 1] + grid[0][j]
dp[i][0] = dp[i - 1][0] + grid[i][0]
1 class Solution { 2 public: 3 int minPathSum(vector<vector<int>>& grid) { 4 int row = grid.size(), col = grid[0].size(); 5 vector<vector<int>> dp(row, vector<int>(col)); 6 7 for (int i = 0; i < row; i++) { 8 for (int j = 0; j < col; j++) { 9 if (!i && !j) { 10 dp[i][j] = grid[i][j]; 11 } 12 else if (!i && j) { 13 dp[i][j] = dp[i][j - 1] + grid[i][j]; 14 } 15 else if (i && !j) { 16 dp[i][j] = dp[i - 1][j] + grid[i][j]; 17 } 18 else { 19 dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; 20 } 21 } 22 } 23 24 return dp[row - 1][col - 1]; 25 } 26 };
3. 在法2的基础上优化空间
1 class Solution { 2 public: 3 int minPathSum(vector<vector<int>>& grid) { 4 int row = grid.size(), col = grid[0].size(); 5 vector<int> dp(col); 6 7 dp[0] = grid[0][0]; 8 for (int i = 0; i < row; i++) { 9 for (int j = 0; j < col; j++) { 10 if (!i && !j) { 11 dp[j] = grid[i][j]; 12 } 13 else if (!i && j) { 14 dp[j] = dp[j - 1] + grid[i][j]; 15 } 16 else if (i && !j) { 17 dp[j] += grid[i][j]; 18 } 19 else { 20 dp[j] = min(dp[j], dp[j - 1]) + grid[i][j]; 21 } 22 } 23 } 24 25 return dp[col - 1]; 26 } 27 };
原文：http://www.cnblogs.com/VickyWang/p/6240818.html