题目描述:
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,使得所有奇数位于数组的前半部分,所有偶数位于数组的后半部分。
示例:
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:
1 <= nums.length <= 50000
1 <= nums[i] <= 10000
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/diao-zheng-shu-zu-shun-xu-shi-qi-shu-wei-yu-ou-shu-qian-mian-lcof
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
解法:
class Solution {
public int[] exchange(int[] nums) {
if(nums == null || nums.length <= 1){
return nums;
}
for (int i = 0,j = 0; i < nums.length&&j < nums.length; ) {
if(nums[j] % 2 == 0){
j++;
}else {
swap(nums,i++,j++);
}
}
return nums;
}
private void swap(int[] nums,int i,int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
优化后解法,来源题解:
class Solution {
public int[] exchange(int[] nums) {
if (nums == null || nums.length <= 1) {
return nums;
}
int left = 0;
int right = nums.length - 1;
while (left < right) {
while (left < right && nums[left] % 2 == 1) {
left++;
}
while (left < right && nums[right] % 2 == 0){
right--;
}
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
left++;
right--;
}
return nums;
}
}
保持数字顺序不变(题目未要求):
class Solution {
public int[] exchange(int[] nums) {
if (nums == null || nums.length <= 1) {
return nums;
}
int index = 0;
for (int i = 0; i < nums.length; i++) {
if(nums[i] %2 == 1){
move(nums,i,index);
index++;
}
}
return nums;
}
private void move(int[] nums,int i,int index) {
int temp = nums[i];
for (int j = i; j > index; j--) {
nums[j] = nums[j-1];
}
nums[index] = temp;
}
}