一、题目

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.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int val = ...; // Value to remove
int[] expectedNums = [...]; // The expected answer with correct length.
                            // It is sorted with no values equaling val.

int k = removeElement(nums, val); // Calls your implementation

assert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
    assert nums[i] == expectedNums[i];
}
If all assertions pass, then your solution will be accepted.

 

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).
Example 2:

Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).
 

Constraints:

0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100

二、 思路

有要原地修改、操作数组的问题考虑快慢指针。通过一个快指针(fast)和一个慢指针(slow)在一个for循环下完成两个for循环的工作。

二、自己写的

1.代码

第一次写

class Solution {
    public int removeElement(int[] nums, int val) {
        int slow = 0, fast = 0;
        while (fast < nums.length) {
            if (nums[fast] != val) {
                slow++;
                nums[slow] = nums[fast];
            }
            fast++;
        }
        return slow;
    }
}

//错误1:瞎几把写,就知道套,没弄懂到底slow和fast代表什么就写了

第二次写

class Solution {
    public int removeElement(int[] nums, int val) {
        //slow里面放的是不等于val的值
        int slow = 0, fast = 0;
        while (fast < nums.length) {
            //slow从0开始,fast从0开始,要判断nums[0]等不等于val,不等于的话赋值然后slow++
            if (nums[fast] != val) {
                nums[slow] = nums[fast];
                slow++;
            }
            fast++;
        }
        //最后的数组长度就是slow
        return slow;
    }
}

//正确,明白了slow里面到底放的是什么

三、标准答案

1.方法一(快慢指针)

class Solution {
    public int removeElement(int[] nums, int val) {
        //slow里面放的是不等于val的值
        int slow = 0, fast = 0;
        while (fast < nums.length) {
            //slow从0开始,fast从0开始,要判断0等不等于val,不等于的话赋值然后slow++
            if (nums[fast] != val) {
                nums[slow] = nums[fast];
                slow++;
            }
            fast++;
        }
        return slow;
    }
}

3.总结

数组的双指针感觉要点就三个,第一个是定义slow和fast的时候,一般就都定义成slow=0,fast=0,但是要搞清楚里面的逻辑,到底是slow里面包含的是什么;第二个点就是明白if里面到底应该怎么判断,slow++放在哪,同样,也是搞清楚逻辑,slow里面到底放的是什么;最后一个就是返回的值是什么,同理还是要明白slow里面到底放的是什么,就能明白了。