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).

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

// 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];
}
};