leetcode 26. Remove Duplicates from Sorted Array

题目

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1（超时，这么笨的办法，我还想了那么久，(⊙﹏⊙)）

 public int removeDuplicates(int[] nums) {        if(nums.length==0||nums.length==1)            return nums.length;        int i=0;        int j=1;        int k=nums.length-1;        while(i<nums.length && j<nums.length && j<k){            if(nums[i]==nums[j])            {                do{                    moveBack(nums,j,k);                    k--;                }while((nums[i]==nums[j])&& j<k);                moveBack(nums,i,k);                k--;            }            else{                i++;                j=i+1;            }        }      return k+1;    }    public void moveBack(int[] nums,int j,int k){        int temp=nums[j];        nums[j]=nums[k];        nums[k]=temp;        String b=Arrays.toString(nums);        System.out.println("Before:"+b);        if(k-j>1)        Arrays.sort(nums, j, k);         b=Arrays.toString(nums);        System.out.println("After:"+b);    }

解2（让元素后面的元素与前面的元素相比）

public class Solution {    public int removeDuplicates(int[] nums) {    if(nums.length ==0){      return 0;    }    int count=1;    for(int i=1; i<nums.length; ++i){      if(nums[i] != nums[i-1]){ //注意这行代码        nums[count]=nums[i];        count++;      }    }    return count;  }}

解3（ 让前面的元素与后面的元素相比）

public int removeDuplicates(int[] nums) {    if(nums.length ==0){      return 0;    }    int count=1;    for(int i=1; i<nums.length; ++i){      if(nums[count-1] != nums[i]){ //注意这行代码        nums[count]=nums[i];        count++;      }    }    return count;  }

参考链接

1：http://www.tuicool.com/articles/ayeuYf