【LeetCode】1217. Play with Chips

• 2019 年 11 月 8 日
• 筆記

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本文鏈接：https://blog.csdn.net/shiliang97/article/details/102527525

There are some chips, and the i-th chip is at position chips[i].

You can perform any of the two following types of moves any number of times (possibly zero) on any chip:

Move the i-th chip by 2 units to the left or to the right with a cost of 0. Move the i-th chip by 1 unit to the left or to the right with a cost of 1. There can be two or more chips at the same position initially.

Return the minimum cost needed to move all the chips to the same position (any position).

Example 1:

Input: chips = [1,2,3] Output: 1 Explanation: Second chip will be moved to positon 3 with cost 1. First chip will be moved to position 3 with cost 0. Total cost is 1. Example 2:

Input: chips = [2,2,2,3,3] Output: 2 Explanation: Both fourth and fifth chip will be moved to position two with cost 1. Total minimum cost will be 2.

來源：力扣（LeetCode） 鏈接：https://leetcode-cn.com/problems/play-with-chips 著作權歸領扣網路所有。商業轉載請聯繫官方授權，非商業轉載請註明出處。

沒想到這道題就是求 奇數或偶數個數的最小值

class Solution {  public:      int minCostToMoveChips(vector<int>& chips) {          int ji=0,ou=0;          for(int i=0;i<chips.size();i++){              if(chips[i]%2==0){                  ou++;              }else{                  ji++;              }          }          if(ji<ou){              ou=ji;          }return ou;      }  };