description: Partition an integers array into odd number first and even number second.
Have you met this question in a real interview? Yes Example Given [1, 2, 3, 4], return [1, 3, 2, 4]
非常簡(jiǎn)單，直接使用two pointers 的算法就能夠非常好的處理。
public class Solution { /** * @param nums: an array of integers * @return: nothing */ public void partitionArray(int[] nums) { // write your code here; if (nums == null || nums.length == 0 || nums.length == 1) { return; } int left = 0; int right = nums.length - 1; while (left <= right) { if (left >= right) { return; } while (left <= right && nums[left] % 2 == 1) { left++; } while (left <= right && nums[right] % 2 == 0) { right--; } if (left <= right) { int temp = nums[left]; nums[left] = nums[right]; nums[right] = temp; left++; right--; } } }}新聞熱點(diǎn)
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