LeetCode -- 120. Triangle

题目：

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：

这道题有些难度，之前第一想法用从上到下的方法寻找最短路径，但是有点傻逼，用了贪心的思想，结果当然不对，没有AC；转念一想，这TM是DP啊，于是又是一顿写，还是不对，边界处理比较麻烦。参考了大佬的想法，从下往上计算路径，一次就AC了，思路很重要，事半功倍~

令 sz=triangle.size();

初始状态：

d[sz−1][i]=triangle[sz−1][i],0≤i≤sz−1;

状态转移方程：

d[i][j]=min(d[i+1][j],d[i+1][j+1])+triangle[i][j],0≤j≤i≤sz−2;

d[0][0]即为最短路径。

C++代码如下：

class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int sz = triangle.size();
vector<vector<int>> d(sz,vector<int>(sz,0));
for(int j=0; j<sz; j++)
{
d[sz-1][j] = triangle[sz-1][j];
}

for(int i = sz-2; i>=0;i--)
{
for(int j=0;j<=i;j++)
{
d[i][j] = min(d[i+1][j], d[i+1][j+1]) + triangle[i][j];
}
}

return d[0][0];
}
};