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.
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 }