LeetCode第26题
Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return 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.
Example 1:
Given nums = [1,1,2],
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:
Given nums = [0,0,1,1,1,2,2,3,3,4],
Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
It doesn't matter what values are set beyond the returned length.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 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]);}翻译：
给定一个已排序的数组，删除其中重复的部分，保证每个数字只出现一次，返回一个新的数组长度。
不要申明额外的数组空间，必须保证算法复杂度为O(1)
思路：
既然不能申明额外的数组，那只能在原来的数组上做变动
变动前：[1,1,2,3,3]
变动后：[1,2,3,3,3]
前3个值[1,2,3]就是我们所需要的
代码：
class Solution { public int removeDuplicates(int[] nums) { if(nums.length == 0) return 0; int j =0; for(int i = 0;i<nums.length;i++){ if(nums[j]!=nums[i]){ j++; nums[j] = nums[i]; } } return j+1; }}原数组遍历一遍后，将不重复的数字保存在数组的前面，j就是我们需要的数据的最大下标，那么j+1就是我们需要的长度
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原始发表：2019-05-06，如有侵权请联系 cloudcommunity@tencent.com 删除
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