Two pointers (5) -- Remove Duplicates from Sorted Array I, II, Move Zeroes, Remove Element

Remove Duplicates from Sorted Array

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 = 

[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 new length.

重复的值是不需要移动的，只需要确保不重复的值放在上一个不重复的值后面

int removeDuplicates(int A[], int n) {
if(n < 2) return n;
int id = 1;
for(int i = 1; i < n; ++i)
if(A[i] != A[i-1]) A[id++] = A[i];
return id;
}

Remove Duplicates from Sorted Array II
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?

和上一题的思路是一样的。只不过移动元素的条件变为数字连续的个数小于3。

int removeDuplicates(vector<int>& nums) {
if (nums.empty()) return 0;
int count = 0;
int index = 1;
int last = nums[0];
for (int i = 1; i < nums.size(); i++){
if (nums[i] == nums[i-1]) count++;
else count = 0;
if (count < 2)
nums[index++] = nums[i];
}
return index;
}

Move Zeroes

Given an array 

nums

, write a function to move all 

0

's
to the end of it while maintaining the relative order of the non-zero elements.

For example, given 

nums = [0, 1, 0, 3, 12]

, after calling your function, 

nums

 should
be 

[1, 3, 12, 0, 0]

.

Note:

You must do this in-place without making a copy of the array.
Minimize the total number of operations.

和上面的思路很像，只关心非零点的位置是否正确；最后用0填充数组末尾部分。

void moveZeroes(vector<int>& nums) {
int j = 0;
// move all the nonzero elements advance
for (int i = 0; i < nums.size(); i++) {
if (nums[i] != 0) {
nums[j++] = nums[i];
}
}
//fill the end of array with zero
for (;j < nums.size(); j++) {
nums[j] = 0;
}
}

Remove Element

Given an array and a value, 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 in place with constant memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example:

Given input array nums = 

[3,2,2,3]

, val = 

3

Your function should return length = 2, with the first two elements of nums being 2.

思路同上。

int removeElement(vector<int>& nums, int val) {
int begin = 0;
for (int i = 0 ; i < nums.size(); i++) {
if(nums[i] != val)
nums[begin++] = nums[i];
}
return begin;
}