剑指 Offer 21. 调整数组顺序使奇数位于偶数前面
本问题对应的 leetcode 原文链接：剑指 Offer 21. 调整数组顺序使奇数位于偶数前面
问题描述
输入一个整数数组，实现一个函数来调整该数组中数字的顺序，使得所有奇数在数组的前半部分，所有偶数在数组的后半部分。
示例 1：
输入：nums = [1,2,3,4]
输出：[1,3,2,4]
注：[3,1,2,4] 也是正确的答案之一。
提示：
0 <= nums.length <= 500000 <= nums[i] <= 10000
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public int[] exchange(int[] nums) {
if(nums == null || nums.length == 0){
return nums;
}
int left = 0;
int right = nums.length - 1;
while(left < right){
while(left < right && nums[left] % 2 != 0) left++;
while(left < right && nums[right] % 2 != 1) right--;
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
}
return nums;
}
}
时间复杂度：O(n)
额外空间复杂度：O(1)
Python
class Solution(object):
def exchange(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
if nums is None or len(nums) == 0:
return nums
left = 0
right = len(nums) - 1
while left < right:
while left < right and nums[left] % 2 != 0:
left += 1
while left < right and nums[right] % 2 != 1:
right -= 1
temp = nums[left]
nums[left] = nums[right]
nums[right] = temp
return nums
C++
class Solution {
public:
vector<int> exchange(vector<int>& nums) {
if (nums.empty()) {
return nums;
}
int left = 0;
int right = nums.size() - 1;
while (left < right) {
while (left < right && nums[left] % 2 != 0) left++;
while (left < right && nums[right] % 2 != 1) right--;
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
}
return nums;
}
};
Go
func exchange(nums []int) []int {
if len(nums) == 0 {
return nums
}
left := 0
right := len(nums) - 1
for left < right {
for left < right && nums[left]%2 != 0 {
left++
}
for left < right && nums[right]%2 != 1 {
right--
}
temp := nums[left]
nums[left] = nums[right]
nums[right] = temp
}
return nums
}
JS
/**
* @param {number[]} nums
* @return {number[]}
*/
var exchange = function(nums) {
if (nums == null || nums.length == 0) {
return nums;
}
let left = 0;
let right = nums.length - 1;
while (left < right) {
while (left < right && nums[left] % 2 !== 0) {
left++;
}
while (left < right && nums[right] % 2 !== 1) {
right--;
}
const temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
}
return nums;
};