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.
class Solution {
public:
int minimumTotal(vector<vector<int> > &triangle)
{
int rows = triangle.size();//该三角形的行和最大列是相等的，所用的空间复杂度即O(rows)
if(rows <= 0)return 0;
vector<int> dp(rows,0);
dp[0] = triangle[0][0];
int i,j;
for (i = 1;i < rows;i++)
{
for (j = triangle[i].size() - 1;j >= 0;--j)
{
if(j == 0)dp[j] = dp[j] + triangle[i][j];//第一列元素
else if(j == triangle[i].size()-1) dp[j] = dp[j-1] + triangle[i][j];//最后一列元素
else dp[j] = min(dp[j-1],dp[j]) + triangle[i][j];//中间的元素。
}
}
int res = dp[0];
for(i = 1;i < rows;i++)
{
if(dp[i] < res)res = dp[i];
}
return res;
}
};原文：http://blog.csdn.net/fangjian1204/article/details/39136623