Remove Duplicates from Sorted Array(II)

• Remove Duplicates from Sorted Array
  • 题目描述
  • 解法
• Remove Duplicates from Sorted Array II
  • 题目描述
  • 解法

Remove Duplicates from Sorted Array

题目描述

Given a sorted array, 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 in place with constant memory.

For example,

Given input array 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 new length.

解法

代码如下:

public static int removeDuplicates( int[] nums ) {
    if( nums.length < 2 ) return nums.length;
    int index = 0;
    for( int i = 1; i < nums.length; ++i )
        if( nums[i] != nums[index] )
            nums[++index] = nums[i];
    return index + 1;
}

思考过程: [0...n-1]数组经过去重之后变为[0...index](index <= n-1), 所以可以使用index变量来维护新数组的元素范围, 在这个过程中遍历数组, 不断加入非重复元素.

时空复杂度: 时间复杂度是O(n), 空间复杂度是O(1)

Remove Duplicates from Sorted Array II

题目描述

Follow up for “Remove Duplicates”: What if duplicates are allowed at most twice?

For example,

Given sorted array nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn’t matter what you leave beyond the new length.

解法

代码如下:

public static int removeDuplicates( int[] nums ) {
    int k = 2;
    if( nums.length < k + 1 ) return nums.length;
    int index = k-1;
    for( int i = k; i < nums.length; ++i )
        if( nums[i] != nums[index-k+1] )
            nums[++index] = nums[i];
    return index + 1;
}

思考过程: 题目要求可以重复2个元素, 这里我把代码修改为可以重复k个元素, 原理与上题一样.

时空复杂度: 时间复杂度是O(n), 空间复杂度是O(1)