LeetCode removeDuplicatesFromSortedArray两种解法方法

// Date : 2016.08.01

// Author : yqtao

// https://github.com/yqtaowhu

/[b]************************************************************************[/b]

*

* 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].

*

*

[b]************************************************************************[/b]/

//solution 1:using stl is very easy.
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.empty()) return 0;
auto cur=nums.begin();
for (;cur!=nums.end()-1;) {
auto next=cur+1;
if (*cur==*next)
nums.erase(cur);
else ++cur;
}
return nums.size();
}
};

//solution 2: 需要注意的是题目中，仅仅说是容器中前面几个数正确即可，因此我们不必要删除每个数
//仅仅是用后一个数去覆盖前一个重复的数即可。
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.empty()) return 0;
int id = 1;
for(int i = 1; i < nums.size(); ++i)
if(nums[i] != nums[i-1])
nums[id++] = nums[i];
return id;
}
};