LeetCode 120. Triangle 题解

120. Triangle

 

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Difficulty: Medium
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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.

解题思路：

设triangle共有n行，找到一个从上到下的最小路径可等效为找到一个从下到上的最短路径。假设输入为一个二维数组（二维vector同理）。令dp[i][j]为从triangle底部到triangle[i][j]的最小值。则有

dp[n-1][j]=triangle[n-1][j];

dp[i][j]=min(dp[i+1][j],dp[i+1][j+1])+triangle[i][j] ;(0<=i<=n-2,0<=j<=i)

由此可得从上到下最小路径为dp[0][0];

代码展示：

class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int row=triangle.size();
int col=row;
vector<int> ans[row];
for(int i=0;i<col;i++)
{
ans[row-1].push_back(triangle[row-1][i]);
}
for(int i=row-2;i>=0;i--)
{
for(int j=0;j<=i;j++)
{
ans[i].push_back(min(ans[i+1][j],ans[i+1][j+1])+triangle[i][j]);
}
}
return ans[0][0];
}
};