[LeetCode] 80. Remove Duplicates from Sorted Array II

轉自LeetCode

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

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

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn't matter what you leave beyond the new length.

<Solution>

這題是 Remove Duplicates from Sorted Array 的衍生題

這次多了可以指定每個數字能重覆幾次

思考方式還是一樣，只是多了一個 counter 來滿足題目需求

code 如下
c++

	class Solution {
	public:
	int removeDuplicates(vector<int>& nums) {
	int newLen = !nums.empty();
	int cnt;
	for(int i = 0; i < nums.size(); i++) {
	if(i == 0) {
	//> first element should be in the array
	cnt = 1;
	}
	else if(nums[i] == nums[newLen - 1] && cnt < 2) {
	//> found duplicates but less than 2 times
	nums[newLen++] = nums[i];
	++cnt;
	}
	else if(nums[i] > nums[newLen - 1]) {
	//> non-duplicates
	nums[newLen++] = nums[i];
	cnt = 1;
	}
	}
	return newLen;
	}
	};

view raw removeDuplicatesFromSortedArrayII.cpp hosted with ❤ by GitHub

kotlin

	class Solution {
	fun removeDuplicates(nums: IntArray): Int {
	var index = 1
	var count = 1
	for (i in 1 until nums.size) {
	if (nums[i] > nums[index-1]) {
	nums[index++] = nums[i]
	count = 1
	} else if (nums[i] == nums[index-1] && count == 1) {
	nums[index++] = nums[i]
	count = 2
	}
	}
	return index
	}
	}

view raw remove_duplicates_from_sorted_array_II.kt hosted with ❤ by GitHub