【leetcode -26】Remove Duplicates from Sorted Array-优快云博客

本文介绍了一种在原地去除排序数组中重复元素的算法，通过一次遍历即可完成，满足O(1)额外内存限制。举例说明了如何将[1,1,2]处理为长度为2的数组，保留前两个唯一元素；同样地，[0,0,1,1,1,2,2,3,3,4]被处理为长度为5的数组，保留前五个唯一元素。

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题目描述

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]);
}

题解如下

由于数组是排序好的，所以只需要相邻之间进行比较判断

代码如下

class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums.length == 0){
            return 0;
        }
        int i=0;
        for(int j=1;j<nums.length;j++){
            if(nums[i] != nums[j]){
                i++;
            }
            nums[i] = nums[j];
        }
        return i+1;
    }
}

疑惑：原本是想利用集合的特性来统计不相同数字的个数的，但不知道为什么行不通。有空再思考下。。。