LintCode: Minimum Path Sum

简介:

C++,

time: O(m*n)

space: O(m*n)

复制代码
 1 class Solution {
 2 public:
 3     /**
 4      * @param grid: a list of lists of integers.
 5      * @return: An integer, minimizes the sum of all numbers along its path
 6      */
 7     int minPathSum(vector<vector<int> > &grid) {
 8         // write your code here
 9         int m = grid.size(), n = grid[0].size();
10         vector<vector<int> > dp(m, vector<int>(n));
11         for (int i = 0; i < m; i++) {
12             for (int j = 0; j < n; j++) {
13                 if (i == 0) {
14                     if (j == 0) {
15                         dp[i][j] = grid[i][j];
16                     } else {
17                         dp[i][j] = dp[i][j-1] + grid[i][j];
18                     }
19                 } else if (j == 0) {
20                     dp[i][j] = dp[i-1][j] + grid[i][j];
21                 } else {
22                     dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
23                 }
24             }
25         }
26         return dp[m-1][n-1];
27     }
28 };
复制代码

C++,

time: O(m*n)

space: O(m)

复制代码
 1 class Solution {
 2 public:
 3     /**
 4      * @param grid: a list of lists of integers.
 5      * @return: An integer, minimizes the sum of all numbers along its path
 6      */
 7     int minPathSum(vector<vector<int> > &grid) {
 8         // write your code here
 9         int m = grid.size(), n = grid[0].size();
10         vector<int> dp(n);
11         for (int i = 0; i < m; i++) {
12             for (int j = 0; j < n; j++) {
13                 if (i == 0) {
14                     if (j == 0) {
15                         dp[j] = grid[i][j];
16                     } else {
17                         dp[j] = dp[j-1] + grid[i][j];
18                     }
19                 } else if (j == 0) {
20                     dp[j] = dp[j] + grid[i][j];
21                 } else {
22                     dp[j] = min(dp[j], dp[j - 1]) + grid[i][j];
23                 }
24             }
25         }
26         return dp[n-1];
27     }
28 };
复制代码

 


本文转自ZH奶酪博客园博客,原文链接:http://www.cnblogs.com/CheeseZH/p/5000605.html,如需转载请自行联系原作者

相关文章
LeetCode 64. Minimum Path Sum
给定m x n网格填充非负数,找到从左上到右下的路径,这最小化了沿其路径的所有数字的总和。 注意:您只能在任何时间点向下或向右移动。
98 0
LeetCode 64. Minimum Path Sum
LeetCode 209. Minimum Size Subarray Sum
给定一个含有 n 个正整数的数组和一个正整数 s ,找出该数组中满足其和 ≥ s 的长度最小的连续子数组。如果不存在符合条件的连续子数组,返回 0。
114 0
LeetCode 209. Minimum Size Subarray Sum
LeetCode 216. Combination Sum III
找出所有相加之和为 n 的 k 个数的组合。组合中只允许含有 1 - 9 的正整数,并且每种组合中不存在重复的数字。
105 0
LeetCode 216. Combination Sum III