题目

We have n chips, where the position of the ith chip is position[i].

We need to move all the chips to the same position. In one step, we can change the position of the ith chip from position[i] to:

  • position[i] + 2 or position[i] - 2 with cost = 0.
  • position[i] + 1 or position[i] - 1 with cost = 1.

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

Example 1:

image.png

  1. Input: position = [1,2,3]
  2. Output: 1
  3. Explanation: First step: Move the chip at position 3 to position 1 with cost = 0.
  4. Second step: Move the chip at position 2 to position 1 with cost = 1.
  5. Total cost is 1.

Example 2:

image.png

  1. Input: position = [2,2,2,3,3]
  2. Output: 2
  3. Explanation: We can move the two chips at poistion 3 to position 2. Each move has cost = 1. The total cost = 2.

Example 3:

  1. Input: position = [1,1000000000]
  2. Output: 1

Constraints:

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

题意

有若干个位置,每个位置上有若干个硬币,现在要将所有硬币移到一个位置。将一个硬币移动一个位置的cost为1,移动两个位置的cost为0,求需要的最小cost。

思路

因为移动两个位置的cost为0,所以所有奇数位的硬币都可以移到位置1,所有偶数位的硬币都可以移到位置2,最后只要考虑从1移到2或从2移到1即可。


代码实现

Java

  1. class Solution {
  2. public int minCostToMoveChips(int[] position) {
  3. int odd = 0, even = 0;
  4. for (int i : position) {
  5. if (i % 2 == 0) {
  6. even++;
  7. } else {
  8. odd++;
  9. }
  10. }
  11. return Math.min(odd, even);
  12. }
  13. }