java清空数组的方法_java初始化数组赋值

java清空数组的方法_java初始化数组赋值 26. Remove Duplicates from Sorted ArrayEasyGiven a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.Clarification:Confused why the returned value is an integer but your answer is an array?Note that the input array is passed in by reference, which means a modification to the input array will be known to the caller as well.Internally you can think of this:// nums is passed in by reference. (i.e., without making a copy)int len = removeDuplicates(nums);// any modification to nums in your function would be known by the caller.// using the length returned by your function, it prints the first len elements.for (int i = 0; i < len; i++) { print(nums[i]);}Example 1:Input: nums = [1,1,2]Output: 2, nums = [1,2]Explanation:Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the returned length.Example 2:Input: nums = [0,0,1,1,1,2,2,3,3,4]Output: 5, nums = [0,1,2,3,4]Explanation:Your function should return length = 5, with the first five elements of nums being modified to0, 1, 2, 3, and4 respectively. It doesn't matter what values are set beyondthe returned length.Constraints:0 <= nums.length <= 3 * 104-104 <= nums[i] <= 104nums is sorted in ascending order.给定一个有序数组，删除重复内容，使每个元素只出现一次，并返回新的长度。

题目要求不要为其他数组分配额外的空间，您必须通过在 O（1）额外的内存中就地修改输入数组来实现这一点。

解题思路：这是一道难度简单的题目，已知条件数组是已排序的，且不能新建数组，这就是告诉我们除了创建数组指针只能改变给定数组来得到结果。

class Solution { public int removeDuplicates(int[] nums) { int n = nums.length; if (n == 0) { return 0; //首先排除边界的case } int counter = 0, i = 1; //定义两个指针，counter代表当前去重的长度，i是遍历的指针，为什么i是从1开始？ 因为起始位置的数组元素已经计算到结果中，所以开始从二个元素遍历 while(i < n) { if (nums[counter] == nums[i]) { //判断当前遍历的数组元素和去重后的最大数组元素是否相同，如果相同则继续遍历，不做任何 *** 作。

i++; } else { //如果不同，则将数组当前遍历的元素替换到去重数组 counter++; //在替换前首先将去重指针后移一位 nums[counter] = nums[i]; i++; //继续遍历 } } return counter + 1; //counter是去重指针的最后一位，所以长度需要加1 }}总结：这类数组去重的题会经常出现在面试环节中（3sum），看到数组去重我们就要想到对数组先进行排序，然后通过一个n的遍历就能解决问题。

另外就是不占用额外空间，只能新建指针和当前给定的数组来 *** 作。