80. Remove Duplicates from Sorted Array II

难度: Medium

题目描述

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

 

Example 1:

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -104 <= nums[i] <= 104
• nums is sorted in non-decreasing order.

解题思路

代码实现

解决方案

java

class Solution {
    public int removeDuplicates(int[] nums) {

        //数组分三部分部分，左侧代表符合条件的元素，中间已处理的数据，右侧代表未处理的元素，leftIndex是左侧最后一个元素的索引，rightIndex是右侧数组的第一个元素，如果leftIndex==0|| (nums[rightIndex]!=nums[leftIndex]||(nums[rightIndex]==nums[leftIndex]&&nums[leftIndex]!=nums[leftIndex-1]))leftIndex++;可以将rightIndex数据复制到leftIndex ,rightIndex++;

        int leftIndex = 0;
        int rightIndex = 1;
        for (int i = rightIndex; i < nums.length; i++) {
            if (leftIndex == 0 || (nums[i] != nums[leftIndex]) || nums[leftIndex] != nums[leftIndex - 1]) {
                leftIndex++;
                nums[leftIndex] = nums[i];

            }
        }
        return leftIndex + 1;

    }
}