[array] leetCode-27. Remove Element

27. Remove Element - Easy

descrition

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 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

Given nums = [3,2,2,3], val = 3,

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

解析

核心思想：双指针；注意需求的分析，根据需求进一步优化设计。

方法 1

参见 code： int removeElementKeepOrder(vector

快慢指针，icur 指向新数组的结尾，i 遍历数组，只要当前值不等于 val 则进行复制。

方法 2

注意到题目的两个前提条件：（1）数组中元素的位置可以改变；（2）超过最新长度 length 后面的元素无关紧要。

这两个条件使得算法进一步优化成为可能。比如当 nums=[1,2,3,5,4],val=4，如果使用方法 1，将产生 4 次赋值操作。而 nums=[4,1,2,3,5],val=4 时，也将产生 4 次赋值操作，实际上我们可以直接将 4 直接删掉的。

还是两个指针，只不过这是两个对立的指针，icur 从前往后，指向当前要检查的数，在此之前的数都是满足要求的，当 nums[icur] 不满足要求时，值需要将 nums[iend] 赋值给它，并将 iend 自减即可， 这相当于之间删除 val。虽然时间复杂度还是 O(n)，但实际上赋值操作的次数只等于需要删除的元素个数，当需要删除的元素个数很少时，算法很高效。

code

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

class Solution{
public:
	int removeElement(vector<int>& nums, int val){
		//return removeElementKeepOrder(nums, val);
		return removeElementChangePlace(nums, val);
	}

	// time-O(n), space-O(1)
	int removeElementKeepOrder(vector<int>& nums, int val){
		int icur = 0;
		for(int i=0; i<nums.size(); i++){
			if(nums[i] != val){
				nums[icur++] = nums[i];
			}
		}

		return icur; // return new length
	}

	// time-O(n). In this approach, the number of assignment operation is 
	// equal to the number of elements to remove. So it is more efficent if 
	// elements to remove are rare.
	// Note: this optimization is based on the condition of 
	//     (1) The order of elements can be changed
	//     (2) It doesn't matter what you leave beyond the new length.   
	// space-O(1),
	int removeElementChangePlace(vector<int>& nums, int val){
		int icur = 0;
		int iend = nums.size() - 1;
		while(icur<=iend){
			if(nums[icur] == val){
				// move nums[icur] to the end and delete which operation is equal to subtract iend.
				// note: icur can't be decrease, because the current element
				// dosen't check 
				nums[icur] = nums[iend];
				iend--;
			}else{
				icur++;
			}
		}

		return iend+1;
	}
};

int main()
{
	return 0;
}