26.Remove Duplicates from Sorted Array

[yunshuipiao]

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]);
}

题目意思：取出排序数组中的重复元素，要求空间在 O(1)， 返回不重复的元素的个数。

注意最后的题目意思，要求返回的 len 长度中就是不重复的元素。

解法

很简单，遍历的过程中，遇到不一样的元素就将其赋值到前面的位置。

直接看代码：

import org.testng.annotations.Test

fun removeDuplicates(nums: IntArray): Int {
    if (nums.isEmpty()) {
        return 0
    }
    var result = 0
    for (index in 0 until nums.size) {
        if (nums[index] != nums[result]) {
            result += 1
            nums[result] = nums[index]
        }
    }
    return result + 1
}


@Test
fun _0026() {
    arrayListOf(intArrayOf(0,0,1,1,1,2,2,3,3,4)).forEach {
        println(removeDuplicates(it))
    }
}