/**********************************************************************************************************
描述
Given a triangle, find the minimum path sum from top to boom. 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]
]
e minimum path sum from top to boom 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 minimum(vector<vector<int>> & triangle){
for(int i = triangle.size() - 2; i >=0; i--){
for(int j = 0;j < triangle[i].size(); j++){
triangle[i][j] = min(triangle[i + 1][j] , triangle[i + 1][j + 1]);
}
}
return triangle[0][0];
}
};
关键是:状态转移方程

参考资料:
LeetCode 题解
本文介绍了一种解决三角形最小路径和问题的高效算法,通过动态规划从下至上更新节点值,最终得到顶部到底部的最小路径和。关键在于状态转移方程的设计,实现仅需O(n)额外空间。
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