标签：
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
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.
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).
Analyse: top to bottom dynamic programming.
Runtime: 22ms
1 class Solution { 2 public: 3 /** 4 * @param triangle: a list of lists of integers. 5 * @return: An integer, minimum path sum. 6 */ 7 int minimumTotal(vector<vector<int> > &triangle) { 8 // write your code here 9 if (triangle.empty() || triangle[0].empty()) return 0; 10 int m = triangle.size(); 11 vector<int> dp(m, INT_MAX); 12 13 dp[0] = triangle[0][0]; 14 for (int i = 1; i < m; i++) { 15 for (int j = i; j > 0; j--) 16 dp[j] = min(dp[j - 1], dp[j]) + triangle[i][j]; 17 dp[0] += triangle[i][0]; 18 } 19 20 int result = INT_MAX; 21 for (int i = 0; i < m; i++) 22 result = min(result, dp[i]); 23 return result; 24 } 25 };
Analyse: Bottom to top dynamic programming.
Runtime: 38ms
1 class Solution { 2 public: 3 /** 4 * @param triangle: a list of lists of integers. 5 * @return: An integer, minimum path sum. 6 */ 7 int minimumTotal(vector<vector<int> > &triangle) { 8 // write your code here 9 if (triangle.empty() || triangle[0].empty()) return 0; 10 int m = triangle.size(); 11 vector<int> dp = triangle.back(); 12 13 for (int i = m - 2; i >= 0; i--) { 14 for (int j = 0; j <= i; j++) { 15 dp[j] = min(dp[j], dp[j + 1]) + triangle[i][j]; 16 } 17 } 18 return dp[0]; 19 } 20 };
标签：
原文地址：http://www.cnblogs.com/amazingzoe/p/5838540.html
