Remove Duplicates from Sorted Array

描述

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 A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

//时间复杂度O（n),空间复杂度O（1）
void PrintArray(int A[], int n){
for (int i = 0; i < n; i++){
cout << A[i] << ends;
}
cout << endl;
}

int RemoveDuplicates(int nums[], int n){
if (n<=0)
{
return 0;
}
else{
PrintArray(nums, n);

int index = 0;
int hashmap=1;
for (int i = 1; i < n; i++){
if (nums[index] != nums[i]){
index++;
nums[index] = nums[i];
}
}

PrintArray(nums, index + 1);
//PrintArray(nums, n);
return index + 1;

}
return 0;
}

Remove Duplicates from Sorted Array II

描述

Follow up for ”Remove Duplicates”: What if duplicates are allowed at most twice?

For example, Given sorted array A = [1,1,1,2,2,3],

Your function should return length = 5, and A is now [1,1,2,2,3]

分析

加一个变量记录一下元素出现的次数即可。这题因为是已经排序的数组，所以一个变量即可解决。如果没有排序的数组，则需要引入一个hashmap来记录出现次数。

//时间复杂度O（n),空间复杂度O（1）
int RemoveDuplicates2(int nums[], int n){
if (n <= 2){
return n;
}
else{
PrintArray(nums, n);

int index = 2;
for (int i = 2; i < n; i++){
if (nums[i] != nums[index - 2]){
nums[index] = nums[i];
index++;  //有区别的
}
}
PrintArray(nums, index);
return index;
}

return 0;
}