[LeetCode]100. Remove Duplicates from Sorted Array II排序数组去重

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.

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解法：同排序数组去重I，本题的一个额外要求是元素最多可以重复两次。因此需要增加一个计数值，记录当前数字出现次数，计数值cnt初始化为1。同时设置一个索引idx初始化为1，记录下一个保留元素应该保存的位置。从数组的第二个元素开始扫描。如果当前元素和前一个元素相同，则++cnt，此时如果cnt>2了，则跳过这个重复值；如果小于2，则nums[idx++]=nums[i]；如果当前元素和前一个元素不相同，则重置cnt=1，且保留下当前元素nums[idx++]=nums[i]。

class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.size() < 3) return nums.size();
int idx = 1, cnt = 1;
for (int i = 1; i < nums.size(); ++i) {
if (nums[i - 1] == nums[i]) {
++cnt;
if (cnt <= 2) nums[idx++] = nums[i];
}
else {
nums[idx++] = nums[i];
cnt = 1;
}
}
return idx;
}
};