标签:path bottom follow ext pat from 自底向上 tom min
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.(Medium)
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.
分析:
经典的动态规划题,可以有自顶向下,自底向上的解法,是自己学动态规划的第一道题,就不分析了,写一个自底向上的解法。
代码:
1 class Solution { 2 public: 3 int minimumTotal(vector<vector<int>>& triangle) { 4 int sz = triangle.size(); 5 if (sz == 0) { 6 return 0; 7 } 8 int dp[sz][sz]; 9 for (int i = 0; i < sz; ++i) { 10 dp[sz - 1][i] = triangle[sz - 1][i]; 11 } 12 for (int i = sz - 2; i >= 0; --i) { 13 for (int j = 0; j <= i; ++j) { 14 dp[i][j] = min(dp[i + 1][j], dp[i + 1][j + 1]) + triangle[i][j]; 15 } 16 } 17 return dp[0][0]; 18 } 19 };
标签:path bottom follow ext pat from 自底向上 tom min
原文地址:http://www.cnblogs.com/wangxiaobao/p/6091170.html