26、题目描述：
Given a sorted array nums, 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 by modifying the input array in-place with O(1) extra memory.
Example 1:
Given nums = [1,1,2],
Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn’t matter what you leave beyond the returned length.
Example 2:
Given nums = [0,0,1,1,1,2,2,3,3,4],
Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
It doesn’t matter what values are set beyond the returned length.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}
解释：
此道题目我采用了两种方法，一种是用了unordered_set（会将重复的数字覆盖掉），另一种用了vector成员函数unique和erase，思路很简单，代码如下：
C++代码：
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
//first solution
unordered_set<int>result;
vector<int>res;
for(int i = 0;i<nums.size();i++)
{
result.insert(nums[i]);
}
nums.clear();
for(auto c:result)
{
nums.push_back(c);
}
sort(nums.begin(),nums.end());
return nums.size();
//secong solution
nums.erase(unique(nums.begin(),nums.end()),nums.end());
return nums.size();
}
};27、题目描述：
Given an array nums and a value val, remove all instances of that value in-place and return the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
The order of elements can be changed. It doesn’t matter what you leave beyond the new length.
Example 1:
Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
It doesn’t matter what you leave beyond the returned length.
Example 2:
Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
Note that the order of those five elements can be arbitrary.
It doesn’t matter what values are set beyond the returned length.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}
解释：
此道题目我才用了两种方法，第一种是采用了不相等就覆盖的方式，第二种是使用了vector成员函数erase和remove，代码如下：
C++代码：
class Solution {
public:
int removeElement(vector<int>& nums, int val)
{
//first solution
int res = 0;
for(int i = 0;i<nums.size();i++)
{
if(nums[i]!=val)
{
nums[res++]=nums[i];
}
}
return res;
//second solution
nums.erase(remove(nums.begin(),nums.end(),val),nums.end());
return nums.size();
}
};