关关的刷题日记96 – Leetcode 120. Triangle- 专知

关关的刷题日记96 – Leetcode 120. Triangle

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关关的刷题日记96 – 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.

给定一个三角形，找到从顶端到低端的最小路径和。每一次可以移动到下一排的相邻的数。能否实现只需要O(n)空间复杂度的算法，n为三角的行数。

思路

对于每一个点来说，以其为端点的路径要么是从左上点下来的，要么是从右上点下来的，所以以这点为端点的路径和sum[i][j]=min(sum[i-1][j-1], sum[i-1][j])+这点值，采用动规的方法来做。

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int re=INT_MAX;
        for(int i=1; i<triangle.size(); ++i)
        {
            for(int j=0; j<triangle[i].size(); ++j)
            {
                if(j==0)
                    triangle[i][j]=triangle[i-1][j]+triangle[i][j];
                else if(j==i)
                    triangle[i][j]=triangle[i-1][j-1]+triangle[i][j];
                else
                    triangle[i][j]=min(triangle[i-1][j-1],triangle[i-1][j])+triangle[i][j];
            }
        }
        for(int i=0; i<triangle[triangle.size()-1].size(); ++i)
            re=min(re,triangle[triangle.size()-1][i]);
        return re<INT_MAX?re:0;
    }
};