Leetcode: Remove Duplicates from Sorted Array
Leetcode: Remove Duplicates from Sorted Array
卡尔曼和玻尔兹曼谁曼
发布于 2019-01-22 16:02:20
发布于 2019-01-22 16:02:20
题目： Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
Do not allocate extra space for another array, you must do this in place with constant memory.
For example, Given input array A = [1,1,2],
Your function should return length = 2, and A is now [1,2].
这道题和上一道题目比较像：Leetcode: Remove Element 都是通过定义一个伪指针，这个指针记录满足要求的数据位置，当前数据满足要求的时候（不用删除的时候）指针移动一位，最后返回这个伪指针的值。
C++参考代码：
class Solution
{
public:
int removeDuplicates(int A[], int n)
{
if (n == 0) return 0;
int pt = 1;
for (int i = 1; i < n; i++)
{
if (A[i - 1] != A[i])
{
A[pt++] = A[i];
}
}
return pt;
}
};C#参考代码：
public class Solution
{
public int RemoveDuplicates(int[] A)
{
if (A.Length == 0) return 0;
int pt = 1;
for (int i = 1; i < A.Length; i++)
{
if (A[i - 1] != A[i]) A[pt++] = A[i];
}
return pt;
}
}Python参考代码：
class Solution:
# @param a list of integers
# @return an integer
def removeDuplicates(self, A):
count = len(A)
if count == 0:
return 0
pt = 1
for i in range(1, count):
if A[i - 1] != A[i]:
A[pt] = A[i]
pt += 1
return ptJava参考代码：
public class Solution {
public int removeDuplicates(int[] A) {
if (A.length == 0) return 0;
int pt = 1;
for (int i = 1; i < A.length; i++) {
if (A[i - 1] != A[i]) {
A[pt++] = A[i];
}
}
return pt;
}
}本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2015年03月19日，如有侵权请联系 cloudcommunity@tencent.com 删除
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