[LeetCode]Remove Duplicates from Sorted Array II

题目描述

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]

.
给定一排好序的数组，将其元素重复多于2的部分删除，返回新数组的长度

解题思路

我们只需分段求解，记录每一段元素的个数，对个数>=2和<2进行讨论，同时使用一个变量lastPos(初始值为0)来记录上一个无重复元素的位置：

Duplicates>=2: A[lastPos] = compareVal; A[lastPos + 1] = compareVal; lastPos += 2;

Duplicates<2: A[lastPos] = compareVal; lastPos += 1;

代码

public static int removeDuplicates(int[] A) {
if (A == null || A.length == 0)
return 0;
if (A.length == 1)
return 1;

int dup = 0;
int lastPos = 0;
int compareVal;
for (int i = 0; i < A.length;) {
dup = 0;
compareVal = A[i];
while (i < A.length && A[i] == compareVal) {
i++;
dup++;
}

if(dup >= 2){
A[lastPos] = compareVal;
A[lastPos + 1] = compareVal;
lastPos += 2;
} else {
A[lastPos] = compareVal;
lastPos += 1;
}
}

return lastPos;
}