[leetcode.com]算法题目 - Triangle

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.

1 class Solution {

2 public:

3     int minimumTotal(vector<vector<int> > &triangle) {

4         // Start typing your C/C++ solution below

5         // DO NOT write int main() function

6         

7     }

8 };

答题模板

思路：假设三角形共有n行，题目中看出第i行共有i个元素。从top到bottom，我们只考虑minimum path的最后一步是停在了这n个元素的哪一个上面。用数组min_sum(k)表示最后一行到达第k个元素的最小路径（min_sum总长度为n），然后找出min_sum(k)中最小的元素即可。注意一下自上而下递推min_sum时候的递推公式，代码如下：

 1 class Solution {

 2 public:

 3     int minimumTotal(vector<vector<int> > &triangle) {

 4         // Start typing your C/C++ solution below

 5         // DO NOT write int main() function

 6         

 7         

 8         int n = triangle.size();

 9         if (0 == n) return 0;

10         if (1 == n) return triangle[0][0];

11         

12         int *min_sum = new int[n];

13         for(int i=0; i<n;i++)

14             min_sum[i] = 0;

15         

16         min_sum[0]=triangle[0][0];

17         for(int i=1;i<n;i++){

18             for(int j=triangle[i].size()-1; j>=0;j--){

19                 if (0==j){

20                     min_sum[j] += triangle[i][0]; 

21                 }else if (triangle[i].size()-1==j){

22                     min_sum[j] = min_sum[j-1]+triangle[i][j];

23                 }else{

24                     min_sum[j] = min(min_sum[j-1], min_sum[j])+triangle[i][j];

25                 }

26             }

27         }

28         

29         int minTotal = min_sum[0];

30         for(int i=1;i<n;i++){

31             minTotal = min(minTotal, min_sum[i]);

32         }

33         

34         delete[] min_sum;

35         return minTotal;

36     }

37     

38     int min(int a, int b){

39         return (a>b?b:a);

40     }

41 };

Answer

注意：一开始的时候内部j那个循环是正着写，后来发现这样有个问题，就是有可能会使用已经变动的过的值去更新下一列。