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Wednesday, January 9, 2013

[LeetCode] Triangle 解题报告



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.
» Solve this problem

[解题思路]

一维DP。逐行扫描,每一个位置能取得的最小sum,是该元素上面两个能取得的最小sum中最小的那一个sum加上自己的值。只需要开一个数组重复利用就行了。

实现的时候,有些繁琐的地方,这个题比较好从下往上扫描。如果从上往下,其中minV的初始值问题就很头疼。


[Code]

1:    int minimumTotal(vector<vector<int> > &triangle) {  
2:      int row = triangle.size();  
3:      if(row ==0) return 0;  
4:      vector<int> minV(triangle[row-1].size());  
5:      for(int i =row-1; i>=0; i--)  
6:      {  
7:        int col = triangle[i].size();  
8:        for(int j =0; j<col; j++)  
9:        {  
10:          if(i == row-1)  
11:          {  
12:            minV[j] = triangle[i][j];  
13:            continue;  
14:          }  
15:          minV[j] = min(minV[j], minV[j+1]) + triangle[i][j];  
16:        }  
17:      }  
18:      return minV[0];  
19:    }  


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