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.
分析:从最小面一层开始往上计算,设dp[i][j]是以第i层j个元素为起点的最小路径和,动态规划方程如下
dp[i][j] = value[i][j] + max{dp[i-1][j], dp[i-1][j+1]}
因为每一层之和它下一层的值有关,因此只需要一个一位数组保存下层的值,代码如下: 本文地址
1 class Solution { 2 public: 3 int minimumTotal(vector<vector<int> > &triangle) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 const int rows = triangle.size(); 7 int dp[rows]; 8 for(int i = 0; i < rows; i++) 9 dp[i] = triangle[rows-1][i]; 10 for(int i = rows-2; i >= 0; i--) 11 for(int j = 0; j <= i; j++) 12 dp[j] = triangle[i][j] + min(dp[j], dp[j+1]); 13 return dp[0]; 14 } 15 };
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