快慢指针/同向指针

• [0,i)的数据代表处理好的数据
• [i,j)的数据是那些处理过但不需要的数据
• [j,array.length)区间的数据为接下来待处理的数据。

以上三个区间的开和闭需要根据题目要求定义，但是要保持一致。
用此方法处理过的数组，处理好的数据相对位置会保持一致。
例如：
[1,2,0,3,0,8]move zeros to the end->[1,2,3,8,0,0],not[8,2,3,1,0,0]or others

通用步骤：

1. Initialize two pointers i and j, usually both equal to 0
2. while j < array.length
   if we need array[j],then we keep it by assigning array[i]=array[j],and move i forward, make it ready at the next position
   otherwise skip it. We do not need to move i since its spot is not fulfilled

举例：

26. 删除有序数组中的重复项

题解：

//leetcode 26.Remove Duplicates From Sorted Array 快慢指针
//图灵星球题解
public int removeDuplicates(int[] arr)
{
	//initialization
	int i = 0, j = 0;
	while (j < arr.size())
	{
		//if not duplicate, keep it, otherwise skip it
		if (i == 0 || arr[j] != arr[i - 1]) arr[i++] = arr[j++];
		else j++;
	}
	//i is now at the length of the new array,[0,i) is what we want
	return i;
}

//labuladong题解
int removeDuplicates(int[] nums)
{
	if (nums.size() == 0) return 0;
	int slow = 0, fast = 0;
	while (fast < nums.size())
	{
		if (nums[fast] != nums[slow])
		{
			slow++;
			//维护nums[0..slow]无重复
			nums[slow] = nums[fast];
		}
		fast++;
	}
	//数组长度为索引+1
	return slow + 1;
}

左右指针/反向指针

[0,i)和(j,array.length)内的数据均为处理好的数据，[i,j]中的数据待处理。
用此方法处理过的数组不会保留原来元素的相对位置

通用步骤：

1. Initialize two pointers i=0,j=array.length -1
2. while i<=j:
   Decide what you should do based on the value of array[i] and array[j]
   Move at least one pointer forward in its direction

举例：

344. 反转字符串

题解：

public char[] reverseString(char[] str)
{
	//initialize
	int i = 0, j = str.length - 1;
	//two pointers opposite direction
	while (i < j)
	{
		//swap str[i] and str[j]
		char tmp = str[i];
		str[i] = str[j];
		str[j] = tmp;
		i++;
		j--;
	}
	return str;
}

//leetcode官方题解
class Solution {
public:
	void reverseString(vector<char>& s) {
		int n = s.size();
		for (int left = 0, right = n - 1; left < right; ++left, --right) {
			swap(s[left], s[right]);
		}
	}
};