Leetcode_c++：Triangle (120)

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).

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算法

O（N^2）
设状态为f(i,j)，表示从从位置(i,j)出发，路径的最小和，则状态转移方程为

f(i,j)=min{f(i,j+1),f(i+1,j+1)}+(i,j)

//原数组上修改

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// LeetCode, Triangle// 时间复杂度O(n^2)，空间复杂度O(1)class Solution {public:    int minimumTotal (vector<vector<int>>& triangle) {        for (int i = triangle.size() - 2; i >= 0; --i)            for (int j = 0; j < i + 1; ++j)                triangle[i][j] += min(triangle[i + 1][j],                        triangle[i + 1][j + 1]);        return triangle [0][0];    }};