【LeetCode】26. Remove Duplicates from Sorted Array

1.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]);
}

2.Solutions

public static int removeDuplicates(int[] nums) {
    int i = nums.length > 0 ? 1 : 0;
    for (int n : nums)
        if (n > nums[i-1])
            nums[i++] = n;
    return i;
}

Question wants us to return the length of new array after removing duplicates and that we don’t care about what we leave beyond new length , hence we can use i to keep track of the position and update the array.

问题要求我们在删除重复项后返回新数组的长度，并且我们不关心超出新长度的内容，因此我们可以使用i来跟踪位置并更新数组。

public int removeDuplicates(int[] nums) {
    int i = 0;
    for(int n : nums)
        if(i < 1 || n > nums[i - 1]) 
            nums[i++] = n;
    return i;
}

（完）