LeetCode 26. Remove Duplicates from Sorted Array
2017-08-15 20:55
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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 nums =
Your function should return length =
It doesn't matter what you leave beyond the new length.
要求:题目要求将数组中重复的数字去掉,且返回去掉后的数组长度。不能使用额外的辅助空间,只能在原数组上操作。
思路:【1】遍历一遍该数组【2】利用 ^ 异或运算的性质:0^n(任何数) = n(任何数), n^n = 0,即如果数组中两个数字相同则异或操作后为0.
以下是AC代码。
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.empty())
return 0;
int tmp = 0;
for(auto it = nums.begin(); it != nums.end(); )
{
tmp ^= *it;
if(0 == tmp && it != nums.begin()) //考虑第一个数字为0的情况,此时不能讲其删除,应该等到下一个元素异或后的结果
{
tmp = *it; //tmp取得当前的元素值
it = nums.erase(it); //将重复数据删除
}
else{
tmp = *it; //tmp取得当前的元素值
++it;
}
}
return nums.size() ;
}
};
Do not allocate extra space for another array, you must do this in place with constant memory.
For example,
Given input array nums =
[1,1,2],
Your function should return length =
2, with the first two elements of nums being
1and
2respectively.
It doesn't matter what you leave beyond the new length.
要求:题目要求将数组中重复的数字去掉,且返回去掉后的数组长度。不能使用额外的辅助空间,只能在原数组上操作。
思路:【1】遍历一遍该数组【2】利用 ^ 异或运算的性质:0^n(任何数) = n(任何数), n^n = 0,即如果数组中两个数字相同则异或操作后为0.
以下是AC代码。
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if(nums.empty())
return 0;
int tmp = 0;
for(auto it = nums.begin(); it != nums.end(); )
{
tmp ^= *it;
if(0 == tmp && it != nums.begin()) //考虑第一个数字为0的情况,此时不能讲其删除,应该等到下一个元素异或后的结果
{
tmp = *it; //tmp取得当前的元素值
it = nums.erase(it); //将重复数据删除
}
else{
tmp = *it; //tmp取得当前的元素值
++it;
}
}
return nums.size() ;
}
};
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