标签：style blog color strong io for
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.
方法一：
数组f，f[i,j] 表示从（0,0）到（i,j）的路径最小和，然后比较最后一行的最小值。top to down方法
1 class Solution { 2 public: 3 // form top to down 4 int minimumTotal(vector<vector<int> > &triangle) { 5 6 size_t rowNum = triangle.size(); 7 if(rowNum == 0) 8 return 0; 9 10 vector<vector<int> >f(rowNum); 11 12 for(size_t i = 0 ; i< rowNum; i++) 13 { 14 f[i].resize(i+1, 0); 15 } 16 17 for(size_t i = 0 ; i< rowNum; i++) 18 { 19 //printVector(triangle[i]); 20 } 21 22 23 f[0][0] = triangle[0][0]; 24 for(size_t i = 1 ; i< rowNum; i++) 25 { 26 for(size_t j = 0; j <= i; j++) 27 { 28 if(j == 0) 29 f[i][j] = f[i-1][j] + triangle[i][j]; 30 else if(j == i) 31 f[i][j] = f[i-1][j-1] + triangle[i][j]; 32 else 33 { 34 int a = f[i-1][j-1]; 35 int b = f[i-1][j]; 36 f[i][j] = min(a, b) + triangle[i][j]; 37 } 38 } 39 } 40 41 int res = INT_MAX; 42 for(size_t i = 0 ; i< rowNum; i++) 43 { 44 // printVector(f[i]); 45 res = min(res, f[rowNum-1][i]); 46 } 47 48 return res; 49 } 50 };
[LeetCode] Triangle,布布扣,bubuko.com
标签：style blog color strong io for
原文地址：http://www.cnblogs.com/diegodu/p/3850660.html
