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.
NSMutableArray *arr = [NSMutableArray arrayWithArray:@[@"1",@"2",@"2",@"2",@"5",@"6",@"6",@"11",@"10"]];
NSInteger duplicatedCount = 0;//统计重复元素个数
NSInteger size = arr.count;
for (NSInteger i = 1; i<size; i++) {
if ([arr[i] integerValue] == [arr[i-1]integerValue]) {
duplicatedCount++;
}
else
{
arr[i-duplicatedCount] = arr[i];
}
}
//最后size - duplicatedCount为元素个数
NSInteger index = 1;
NSInteger size = arr.count;
NSInteger count = size; //count为不重复元素个数
for (NSInteger i = 1; i<size; i++) {
if ([arr[i] integerValue] != [arr[i-1]integerValue]) {
arr[index] = arr[i];
index++;
}
else
{
count --;
}
}
int curInsertIndex = 0;//当前需要替换元素的位置,并且最后就是新数组的长度
for (int i = 0; i < arr.count; i++)
{
if (curInsertIndex == 0 || [arr[i]integerValue] != [arr[i-1]integerValue])
{
arr[curInsertIndex++] = arr[i];
}
}
