Code & Fun

第84天。

All right,最终还是在期末考的时候断了。

今天的题目是Remove Duplicates from Sorted Array:

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 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 new length.

明明是道水题,但是还是做了挺久的,而且效率也不高:

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int removeDuplicates1(vector<int>& nums) {
int size = nums.size();
for(int i = size - 1;i >= 1;i--) {
if (nums[i] == nums[i-1]) {
swap(nums[i],nums[--size]);
}
}
sort(nums.begin(),nums.begin() + size);
return size;
}

说他效率不高的原因就在于最后要做一次排序。

然后是dicuss中给出的O(n)的解法:

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int removeDuplicates(vector<int>& nums) {
if (nums.size() <= 1) return nums.size();
int end = 1;
for(int i = 1;i < nums.size();i++) {
if (nums[i] != nums[i-1]) nums[end++] = nums[i];
}
return end;
}

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