Question:

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.

Thinking:

  • Method1: Save a temp value to compare. 
      class Solution {
      public int removeDuplicates(int[] nums) {
          int result = 1;
          int len = nums.length;
          if(0 == len)    return 0;
          int temp = nums[0];
          for(int i = 1; i < len; i++){
              int cur = nums[i];
              if(temp != cur){
                  nums[result++] = cur;
                  temp = cur;
              }
          }
          return result;
      }
  }

二刷

和一刷的时候想法相同，使用index控制填入数据的位置。 只使用O（1）额外空间和O（N）时间。

class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int index = 1;
        int pre = nums[0];
        for(int i = 1; i < nums.length; i++){
            if(nums[i] == pre) continue;
            else{
                nums[index++] = nums[i];
                pre = nums[i];
            }
        }
        return index;
    }
}