关关的刷题日记96 – Leetcode 120. Triangle
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。
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;
}
};
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