Leetcode Hot 100刷题记录 -Day3(双指针)
移动零
问题描述:
给定一个数组
nums
,编写一个函数将所有0
移动到数组的末尾,同时保持非零元素的相对顺序。请注意 ,必须在不复制数组的情况下原地对数组进行操作。示例 1:
输入: nums = [0,1,0,3,12] 输出: [1,3,12,0,0]示例 2:
输入: nums = [0] 输出: [0]
//解法一:
class Solution {
public int[] moveZeroes(int[] nums){
int n = nums.length;
int j =0; //指针j为指向已经处理好的序列尾部
//指针i指向未处理好的序列头部
for (int i = 0;i < n; i++){
if(nums[i] != 0){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
j++;
}
}
return nums;
}
}
//解法二:
class Solution {
public int[] moveZeroes(int[] nums){
int n = nums.length;
int j =0;
for (int i = 0;i < n; i++){
if(nums[i] != 0){
nums[j] = nums[i];//指针j指向为0的数,直接将其变为非0的数
j++;
}
}
//第一次循环将前面所有不为0的数全部移动到前面,且指针j最后指向的是已经处理好的序列尾部
//所以指针i和指针j之间的数为0的个数
for (int i = j; i<n; i++){
nums[i] = 0;
}
return nums;
}
}
//带有输入输出:
//解法一:
public class hot4_moveZeroes {
public int[] moveZeroes(int[] nums){
int n = nums.length;
int j =0;
for (int i = 0;i < n; i++){
if(nums[i] != 0){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
j++;
}
}
return nums;
}
public static void main(String[] args) {
hot4_moveZeroes hot4MoveZeroes = new hot4_moveZeroes();
int[] nums = {0,1,0,3,12};
System.out.println("输入: = " + Arrays.toString(nums));
int[] result = hot4MoveZeroes.moveZeroes(nums);
System.out.println("输出:" + Arrays.toString(result));
}
}
//解法二:
public class hot4_moveZeroes {
public int[] moveZeroes(int[] nums){
int n = nums.length;
int j =0;
for (int i = 0;i < n; i++){
if(nums[i] != 0){
nums[j] = nums[i];
j++;
}
}
for (int i = j; i<n; i++){
nums[i] = 0;
}
return nums;
}
public static void main(String[] args) {
hot4_moveZeroes hot4MoveZeroes = new hot4_moveZeroes();
int[] nums = {0,1,0,3,12};
System.out.println("输入: = " + Arrays.toString(nums));
int[] result = hot4MoveZeroes.moveZeroes(nums);
System.out.println("输出:" + Arrays.toString(result));
}
}
知识点总结:
- 方法二的思想