"Leetcode" Array -6 (561)-array Partition I (more abstract topic)

Title Description: Two words are thought-provoking ah ....

Given an array of 2n integers, your task was to group these integers into n pairs of Integer, say (A1, B1), (A2, B2), ..., (an, bn) which makes sum of min (AI, bi) for all I from 1 to n as large as possible.

The general meaning of the feeling topic is to divide the array into a number of two-dollar arrays, summing them in a way that minimizes the value, making and maximizing.

Ideas:

At the beginning of the idea, it is not known right now, is to sort the array, using a pointer to traverse backward , to find the smallest and sum. "Note" The pointer accumulates 2 at a time

"Correct code" once write to ~

1 classSolution {2      Public intArraypairsum (int[] nums) {3         if(nums.length% 2! = 0 | | nums = =NULL) {4             return-1;5         }6 Arrays.sort (nums);7         intMaxsum = 0;8          for(inti = 0; i < nums.length-1;i + = 2) {9Maxsum + = Math.min (Nums[i], nums[i + 1]);Ten         } One         returnmaxsum; A     } -}

Complexity of Time: O (N*LOGN)

Space complexity: O (N*LOGN)

"Leetcode" Array -6 (561)-array Partition I (more abstract topic)