LeetCode(26) Remove Duplicates from Sorted Array

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.

第一次的代码：

public int removeDuplicates(int[] nums) {
if(nums.length==0)
return 0;
else{
int flag=nums[0];
int loc=0;
for(int i=1;i<nums.length;i++){
if(nums[i]!=flag){
flag=nums[i];
}else{
/*需改进的地方*/
for(int j=i;j<nums.length-1;j++){
nums[j]=nums[j+1];
}
i--;
}
loc=i;
if(nums[i]==nums[nums.length-1])
break;
}
return loc+1 ;
}
}

第二次的代码：（类似有两个指针）

public int removeDuplicates(int[] nums){
if(nums.length==0)
return 0;
int loc1=1;
int loc2=0;
while(loc1!=nums.length){
if(nums[loc1]!=nums[loc2]){
loc2++;
nums[loc2]=nums[loc1];
loc1++;
}else{
loc1++;
}
}
return loc2+1;
}

第三次的代码：（简洁代码）

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