标签：style blog color 使用 io strong for ar
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：依次对每一层进行求解，使用数组记录当前层和上一层的结果。对于triangle[i][j]，其上层节点为triangle[i-1][j]和triangle[i-1][j+1]。
1 class Solution { 2 public: 3 int minimumTotal( vector<vector<int>> &triangle ) { 4 if( triangle.empty() ) { return 0; } 5 vector<vector<int>> totals( 2, vector<int>( triangle.size(), 0 ) ); 6 totals[0][0] = triangle[0][0]; 7 for( int i = 1; i < (int)triangle.size(); ++i ) { 8 vector<int> &prev = totals[(i-1)%2], &curr = totals[i%2]; 9 curr[0] = triangle[i][0] + prev[0]; 10 curr[i] = triangle[i][i] + prev[i-1]; 11 for( int k = 1; k < i; ++k ) { 12 curr[k] = triangle[i][k] + min( prev[k-1], prev[k] ); 13 } 14 } 15 vector<int> &curr = totals[((int)triangle.size()-1)%2]; 16 int minTotal = curr[0]; 17 for( size_t i = 1; i != curr.size(); ++i ) { 18 if( minTotal > curr[i] ) { minTotal = curr[i]; } 19 } 20 return minTotal; 21 } 22 };
上述方法采用的是自顶向下的求解策略，若采用自底向上的策略，我们可以获得更加简洁的代码：
1 class Solution { 2 public: 3 int minimumTotal( vector<vector<int>> &triangle ) { 4 if( triangle.empty() ) { return 0; } 5 vector<int> sum = triangle.back(); 6 for( size_t i = triangle.size()-1; i != 0; --i ) { 7 for( size_t k = 0; k != i; ++k ) { 8 sum[k] = triangle[i-1][k] + min( sum[k], sum[k+1] ); 9 } 10 } 11 return sum[0]; 12 } 13 };
标签：style blog color 使用 io strong for ar
原文地址：http://www.cnblogs.com/moderate-fish/p/3913423.html
