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.
这道题目是[url=http://kickcode.iteye.com/blog/2274174l]Remove Duplicates from Sorted Array [/url]的follow up。这里每个元素可以出现两次。我们只需要维护一个计数指针就可以了。当计数指针大于2的时候我们就需要忽略当前的元素,继续往下遍历。代码如下:
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.
这道题目是[url=http://kickcode.iteye.com/blog/2274174l]Remove Duplicates from Sorted Array [/url]的follow up。这里每个元素可以出现两次。我们只需要维护一个计数指针就可以了。当计数指针大于2的时候我们就需要忽略当前的元素,继续往下遍历。代码如下:
public class Solution {
public int removeDuplicates(int[] nums) {
if(nums == null || nums.length == 0) return 0;
int index = 0;
int counter = 0;
for(int i = 0; i < nums.length; i++) {
if(i > 0 && nums[i] == nums[i - 1]) {
if(counter >= 2)
continue;
else {
nums[index ++] = nums[i];
counter ++;
}
} else {
nums[index ++] = nums[i];
counter = 1;
}
}
return index;
}
}