Leetcode-283. Move Zeroes
阿新 • • 發佈:2019-04-29
給定 輸出 != 排列 clas 遍歷 空間 pub problem
地址:https://leetcode.com/problems/move-zeroes/
項目描述:
Given an array nums
, write a function to move all 0
‘s to the end of it while maintaining the relative order of the non-zero elements.
給定一個數組 nums,寫一個函數讓數組中所有為0的元素移到數組的末尾,並且保持非0元素的相對位置
示例:
輸入: [0,1,0,3,12]
輸出: [1,3,12,0,0]
Note:
- You must do this in-place without making a copy of the array.
- Minimize the total number of operations.
方法1:
/**
* 時間復雜度:O(n)
*空間復雜度:O(n)
* @param nums
*/
public void moveZeroes_1(int[] nums) {
ArrayList<Integer> list = new ArrayList<>(); int size = nums.length; for (int i = 0; i < size; i++) { if(nums[i] != 0) { list.add(nums[i]); } } for (int i = 0; i < list.size(); i++) { nums[i] = list.get(i); } for (int i = list.size(); i < size; i++) { nums[i] = 0; } }
方法2:
/**
* 時間復雜度:O(n)* 空間復雜度:O(1)
* @param nums
*/
public void moveZeroes_2(int[] nums) {
int k = 0; int size = nums.length; for (int i = 0; i < size; i++) { if (nums[i] == 0) { swap(nums, i, k++); } } } private void swap(int[] nums, int i, int k) { int temp = nums[k]; nums[k] = nums[i]; nums[i] = temp; }
方法3:
/**
* 時間復雜度:O(n)
* 空間復雜度:O(1)
* @param nums
*/
public static void moveZeroes_3(int[] nums) {
int k = 0; // nums 中,[0...k)的元素均為非0元素
int size = nums.length;
// 遍歷到第 i 個元素後,保證[0...i]中所有非0元素都按照順序排列在[0...k)中
for (int i = 0; i < size; i++) {
if (nums[i] != 0) {
nums[k++] = nums[i];
}
}
// 將 nums 剩余的位置放置為0
for (int i = k; i < size; i++) {
nums[i] = 0;
}
}
Leetcode-283. Move Zeroes