数组02–移除元素

题干

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The relative order of the elements may be changed.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

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:

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,_,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).

暴力解法

// 暴力解法：两个循环，一个for遍历数组，一个for更新数组
// 时间 O(n^2)
// 空间 O(1)

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int size = nums.size();
        for (int i = 0; i < size; i++) {
            if (nums[i] == val){
                for(int j = i + 1; j < size; j++){
                    nums[j-1] = nums[j];
                }
                i --;//因为i后面的都前移一位，所以i也要前移一位，
                     //i++回来，再从这个位置计算，因为新的i位置可能还是那个val，要重新判断
                size --;
            }
        }
        return size;      
    }
};

双指针法

双指针法（快慢指针法）在数组和链表的操作中是非常常见的，很多考察数组、链表、字符串等操作的面试题，都使用双指针法。

双指针法（快慢指针法）： 通过一个快指针和慢指针在一个for循环下完成两个for循环的工作。

定义快慢指针:

• 快指针：寻找新数组的元素 ，新数组就是不含有目标元素的数组
• 慢指针：指向更新新数组下标的位置

很多同学这道题目做的很懵，就是不理解 快慢指针究竟都是什么含义，所以一定要明确含义，后面的思路就更容易理解了。

// 快慢指针法
// 时间O(n)
// 空间O(1)

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int slowIndex = 0;
        for(int fastIndex = 0; fastIndex < nums.size(); fastIndex++){
            if(val != nums[fastIndex]){
                nums[slowIndex] = nums[fastIndex];
                slowIndex++;
            }
        }
        return slowIndex;        
    }
};

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