1、題目

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.

2、代碼實(shí)現(xiàn)

public class Solution {public int removeDuplicates(int[] nums) {if (null == nums || nums.length == 0) {return 0;}int length = nums.length;int newLen = 1;int count = 1;for (int i = 1; i < length; ++i) {if (nums[i] == nums[i - 1]) {count++;if (count <= 2) {nums[newLen++] = nums[i];} } else {count = 1;nums[newLen++] = nums[i];}}return newLen;} }
?

3、總結(jié)

既然看到連續(xù)相等，那我們就要想到前后兩個(gè)不相等作為條件，然后加上通過(guò)一個(gè)變量不超過(guò)2個(gè)就可以了

?

總結(jié)

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