标签:style class blog code ext color
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 onlyO(n) extra space, wheren is the total number of
rows in the triangle.
解题思路:
首先dp是非常easy想到的,用dp[i][j]表示到第i层j列的最小值,那么状态转移为:
dp[i][j]=min(dp[i-1][j],dp[i-1][j-1])+triangle[i][j] ( 0 < j < triangle[i].size() - 1).因为状态转移仅仅与i和i-1
行有关,所以用滚动数组能将其内存优化为O(n)(此处有常系数2).关于内存的优化,事实上是能够做
到严格的O(n)的,我们通过反向更新就可以实现.
解题代码:
class Solution { public: int minimumTotal(vector<vector<int> > &triangle) { const int n = triangle.size(); int dp[n]; dp[0] = triangle[0][0]; for (int i = 1; i < n; ++i) { for (int j = triangle[i].size() - 1; j >= 0 ; --j) { if (j == triangle[i].size() - 1 || !j) dp[j] = (!j ? dp[0] : dp[j-1]) + triangle[i][j]; else dp[j] = min(dp[j],dp[j-1]) + triangle[i][j]; } } return *min_element(dp,dp+n); } };
更简单的代码:
class Solution { public: int minimumTotal(vector<vector<int> > &triangle) { const int n = triangle.size(); int dp[n]; for (int i = n -1; i >= 0; --i) dp[i] = triangle[n-1][i]; for (int i = n -2; i >= 0; --i) for (int j = 0; j < triangle[i].size(); ++j) dp[j] = min(dp[j],dp[j+1]) + triangle[i][j]; return dp[0]; } };
LeetCode:Triangle,布布扣,bubuko.com
标签:style class blog code ext color
原文地址:http://www.cnblogs.com/mfrbuaa/p/3810000.html