Leetcode: Triangle

Leetcode: Triangle

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 is11(I. e., 2 + 3 + 5 + 1 = 11 ).

Note:
Bonus Point if you are able to do this using onlyO(N) Extra space, whereNIs the total number of rows in the triangle.

Address: https://oj.leetcode.com/problems/triangle/

Algorithm: open two n-sized arrays. Suppose we have completed the search for the first line, the minimum path of the End Node with the J element of row I is stored in an array. Assume that the array points to the destination node with the pointer pre. Next, we need to complete the search on line I + 1, the path with the J element of line I + 1 as the final node must be from the J node of line I or the J + 1 node of line I, when calculating the minimum path, you only need to consider these two paths and store the results in another array. This array points to the pointer p. Because only the last result is needed in each search process, we only use two arrays and use the pre and P pointers to reuse these two arrays, in this way, the space requirements of O (n) can be met. Code:

 1 class Solution { 2 public: 3     int minimumTotal(vector<vector<int> > &triangle) { 4         int n = triangle.size(); 5         if(n < 1)    return 0; 6         if(n == 1)   return triangle[0][0]; 7         vector<int> min1(n); 8         vector<int> min2(n); 9         min1[0] = triangle[0][0];10         vector<int> *pre = &min1;11         vector<int> *p   = &min2;12         for(int i = 1; i < n; ++i){13             for(int j = 0; j <= i; ++j){14                 if(j == 0){15                     (*p)[j] = (*pre)[j] + triangle[i][j];16                 }else if(j == i){17                     (*p)[j] = (*pre)[j-1] + triangle[i][j];18                 }else{19                     (*p)[j] = ( (*pre)[j] > (*pre)[j-1] ? (*pre)[j-1] : (*pre)[j] ) + triangle[i][j];20                 }21             }22             vector<int> *t = pre;23             pre = p;24             p = t;25         }26         int result = (*pre)[0];27         for(int i = 1; i < n; ++i){28             if (result > (*pre)[i]) result = (*pre)[i];29         }30         return result;31     }32     33 };

 

Leetcode: Triangle