原题地址:https://oj.leetcode.com/problems/remove-duplicates-from-sorted-array-ii/
题意:
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array A = [1,1,1,2,2,3]
,
Your function should return length = 5
, and A is now [1,1,2,2,3]
.
解题思路:一种巧妙的解法。使用两个指针prev和curr,判断A[curr]是否和A[prev]、A[prev-1]相等,如果相等curr指针继续向后遍历,直到不相等时,将curr指针指向的值赋值给A[prev+1],这样多余的数就都被交换到后面去了。最后prev+1值就是数组的长度。
代码:
class Solution: # @param A a list of integers # @return an integer # @it's a good solution! def removeDuplicates(self, A): if len(A) <= 2: return len(A) prev = 1; curr = 2 while curr < len(A): if A[curr] == A[prev] and A[curr] == A[prev - 1]: curr += 1 else: prev += 1 A[prev] = A[curr] curr += 1 return prev + 1