LeetCode 26.Remove Duplicates from Sorted Array 移除有序数组的重复元素

26.移除有序数组的重复元素

26. Remove Duplicates from Sorted Array

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]);
}

双指针法

这里使用的是快慢指针，从声明中可以发现，要求我们返回的是数组的长度，并且通过这个函数来修改原数组。因此可以使用双指针的方法。当遇到重复的元素的时候，使用快指针的元素进行填充。

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