26. Remove Duplicates from Sorted Array

Question

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. 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.

Solution

由于数组本身已经排序，只要比较当前nums中的元素是否大于上一个保存的数值就可以决定是否保留。
创建一个k记录遍历的位置，每次比较nums[k]与nums[i]的位置元素的大小，如果当前的nums[i]大于nums[k]，则将k位置向后移动1，并将下一个位置记录为nums[i]。

最后返回k+1。

Code

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