Triangle

寻找最小路径和：使用动态规划解决三角形问题

最新推荐文章于 2024-08-16 07:40:36 发布

转载最新推荐文章于 2024-08-16 07:40:36 发布 · 56 阅读

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原文链接：http://www.cnblogs.com/waruzhi/p/3413319.html

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

思路：

经典的DP。因为限制用O（n）的空间，只需保存最下面一行的数字，然后逐行相加即可。

代码：

 1     int min(int a, int b){
 2         if(a < b)
 3             return a;
 4         return b;
 5     }
 6     int minimumTotal(vector<vector<int> > &triangle) {
 7         // IMPORTANT: Please reset any member data you declared, as
 8         // the same Solution instance will be reused for each test case.
 9         int n = triangle.size();
10         if(n == 0)
11             return 0;
12         vector<int> v = triangle[n-1];
13         int m = n;
14         while(m > 1){
15             for(int i = 0; i < m-1; i++){
16                 v[i] = triangle[m-2][i] + min(v[i], v[i+1]);
17             }
18             m--;
19         }
20         return v[0];
21     }