1217. Play with Chips

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.

Constraints:

• 1 <= chips.length <= 100
• 1 <= chips[i] <= 10^9

class Solution {
    public int minCostToMoveChips(int[] chips) {
        int even = 0, odd = 0;
        for(int i = 0; i < chips.length; i++){
            if(chips[i] % 2 == 0) even++;
            else odd++;
        }
        return Math.min(even, odd);
    }
}

 我吐了，谁出的题，晚饭和老八梁非凡一桌。

大意：移动even步不要钱，odd步要钱，所以分别计算chip中odd和even数量取小的一个即可。

1217. Play with Chips

原文：https://www.cnblogs.com/wentiliangkaihua/p/12208417.html