题目
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.
For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0]
给定一个数组,写一个函数,将数组中的所有0挪到数组的末尾,而维持其他所有非0元素的相对位置.举例 input:[0,1,0,3,12] output:[1,3,12,0,0]
代码
方法一：
public static void moveZeroes(int[] nums) {
int[] nonZeroElements = new int[nums.length];
int count = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != 0) {
nonZeroElements[count] = nums[i];
count++;
}
}
for (int i = count; i < nums.length; i++) {
nonZeroElements[count] = 0;
count++;
}
for (int i = 0; i < nums.length; i++) {
nums[i] = nonZeroElements[i];
}
}
方法二：
public static void moveZeroes(int[] nums) {
int size = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > 0) {
nums[size] = nums[i];
size++;
}
}
for (int i = size; i < nums.length; i++) {
nums[i] = 0;
}
}
方法三：
public static void moveZeroes3(int[] nums) {
int k = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > 0) {
if (i != k) { //如果数组中没有0,就不需要交换
swap(nums, i, k++);
} else {
k++;
}
}
}
}
public static void swap(int[] arr, int i, int j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
