Given a sorted array nums, remove the duplicates in-place such that each element appears 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：
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.
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]);
}
解釋下題目：
返回數(shù)組中不重復的數(shù)字的個數(shù)。上面解釋了一下為什么要返回不同數(shù)組的個數(shù)而不是返回處理過后的數(shù)組受葛。不能使用額外的空間，也就是說必須在這個數(shù)組上進行處理鱼喉。
1. 處理之后放入數(shù)組前端
實際耗時：12ms
public int removeDuplicates(int[] nums) {
int length = nums.length;
if (length == 0) {
return 0;
}
int count = 0;
for (int i = 1; i < length; i++) {
if (nums[i] != nums[count]) {
count++;
nums[count] = nums[i];
}
}
return count + 1;
}
??思路：因為是已經(jīng)排過序的數(shù)組墓造，所以相同的數(shù)肯定是相鄰的，而且是刪除重復的（其實也不是刪除）元素，所以最壞的情況是這個數(shù)組本身就沒有重復的纱控。思路就是處理過后就覆蓋掉辆毡，因為肯定有處理后的數(shù)組<=處理前的數(shù)組，所以不用擔心甜害。
