| Excerpt |
|---|
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:
| Code Block |
|---|
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. |
...
| Code Block |
|---|
// 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]);
} |
...
Solution in C++
| Code Block |
|---|
class Solution { public: int removeDuplicates(vector<int>& nums) { map<int,int> mp; for(int i=0; i<nums.size(); i++) { if (!mp[nums[i]]) { mp[nums[i]]=i; nums[ mp.size()-1] = nums[i]; } } return mp.size(); } }; |
Solution in Java
| Code Block |
|---|
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;
} |