leetcodeOJ 120. 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：对空间复杂度有要求，申请一个n大小的数组，初始化为第n-1行的元素，

本题可用动态规划来解，从下往上处理，第i行第j列元素的值为（min（第i+1行第j列元素，第i+1行第j+1列元素）+  第i行第j列元素）

代码如下：

//动态规划
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int n = triangle.size();
int ans
;
for(int i = 0; i < n; i++){
ans[i] = triangle[n-1][i];//初始化为最后一列
}
for(int i = 0; i < n-1; i++){
for(int j = 0; j < n-1-i; j++){
ans[j] = triangle[n-2-i][j] + min(ans[j], ans[j+1]);
}
}
return ans[0];
}
};