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  • LeetCode:Triangle

    题目链接

    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 };

    【版权声明】转载请注明出处:http://www.cnblogs.com/TenosDoIt/p/3436532.html

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  • 原文地址:https://www.cnblogs.com/TenosDoIt/p/3436532.html
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