# 删除排序数组中的重复项 Remove Duplicates from Sorted Array

## 题目

给定一个排序数组，你需要在原地删除重复出现的元素，使得每个元素只出现一次，返回移除后数组的新长度>

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.
```

**Clarification:**

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

```
// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}
```

## 思路&题解

```
public int removeDuplicates(int[] nums) {
if (nums.length <= 1) {
return nums.length;
}
int len = 1;
int end = nums.length;
for (int i = 0; i < end; i++) {
int j = i + 1;
if (j < nums.length) {
if (nums[i] == nums[j]) {
end--;
while (j < nums.length - 1) {
nums[j] = nums[j + 1];
j++;
}
nums[nums.length - 1] = nums[i];
} else if (nums[i] > nums[j]) {
len = i+1;
break;
}
}else{
len = i+1;
}
}
return len;
}
```

正确

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