以下三个问题的典型的两个指针处理数组的问题，一个指针用于遍历，一个指针用于指向当前处理到位置
一：Remove Element
Given an array and a value, remove all instances of that value in place and return the new length.
The order of elements can be changed. It doesn‘t matter what you leave beyond the new length.
class Solution { public: int removeElement(vector<int>& nums, int val) { int numsSize = nums.size(); int removeStart = numsSize; int i=0; while(i<removeStart){ if(nums[i] == val) { swap(nums[i],nums[--removeStart]); }else{ i++; } } for(int i=numsSize-1;i>=removeStart;i--){ nums.erase(nums.begin()+i); } return removeStart; } };
二：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.
class Solution { public: int removeDuplicates(vector<int>& nums) { int numsSize = nums.size(); int j=0; int repeat = 0; for(int i=0;i<numsSize;i++){ if(i==0){ j=1; repeat = nums[i]; }else{ if(nums[i]!=repeat){ nums[j++] = nums[i]; repeat = nums[i]; } } } for(int i=numsSize-1;i>=j;i--){ nums.erase(nums.begin()+i); } return j; } };
三：Remove Duplicates from Sorted Array II
Follow up for "Remove Duplicates":
What if duplicates are allowed at most twice?
For example,
Given sorted array nums = [1,1,1,2,2,3],
Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn‘t matter what you leave beyond the new length.
class Solution { public: int removeDuplicates(vector<int>& nums) { int numsSize = nums.size(); int j=0; int repeat = 0; int repeatTime = 0; for(int i=0;i<numsSize;i++){ if(i==0){ j=1; repeat = nums[i]; repeatTime = 1; }else{ if(nums[i]!=repeat){ nums[j++]=nums[i]; repeat = nums[i]; repeatTime = 1; }else if(repeatTime < 2){ nums[j++]=nums[i]; repeatTime ++; } } } for(int i=numsSize-1;i>=j;i--){ nums.erase(nums.begin()+i); } return j; } };
Remove Element,Remove Duplicates from Sorted Array,Remove Duplicates from Sorted Array II
原文：http://www.cnblogs.com/zengzy/p/4960608.html