【LeetCode】27. Remove Element - Java实现

1. 题目描述：

Given an array nums and a value val, 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 1:

Given nums = [3,2,2,3], val = 3,
Your function should return length = 2, with the first two elements of nums being 2.
It doesn’t matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,
Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.
Note that the order of those five elements can be arbitrary.
It doesn’t matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeElement(nums, val);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

2. 思路分析：

题目的意思是给定一个数组和一个数值（val），将数组中值为val的元素删除，最后返回非重复数字个数(length)，并且要求不要使用额外的存储空间，且最后数组的前length个数是去掉val的。

我们可以使用快慢指针来记录遍历的坐标，最开始时两个指针都指向第一个数字，如果快指针指向的数字等于val，则快指针向前走一步，如果不等于，则快指针向前走一步，并将快指针指向的数交换到慢指针指向的元素同时慢指针向前走一步，这样当快指针走完整个数组后，慢指针当前的坐标就是不等于val数字的个数。

3. Java代码：

源代码：见我GiHub主页

代码：

public static int removeElement(int[] nums, int val) {
    if (nums.length == 0) {
        return 0;
    }

    int preIndex = 0;
    for (int i = 0; i < nums.length; i++) {
        if (nums[i] != val) {
            nums[preIndex++] = nums[i];
        }
    }
    return preIndex;
}