LeetCode 27. Remove Element

分析

难度 易

来源

https://leetcode.com/problems/remove-element/description/

题目

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]);

} 

 

解答

package LeetCode;

public class L27_RemoveElement {
    public int removeElement(int[] nums, int val) {
        /*
        返回移除val后的数组长度
        同时需要对nums进行修改，使得nums前len个数不包含val值
        前len个数的顺序不做要求
        考虑类似快排的交换方式
         */
        int len=nums.length;
        int start=0,end=len-1;
        while(end>=start){
            //从后往前比较
            while(end>=start&&nums[end]==val)  //如果当前值等于val，比较下一个，直到有不等于val的交换位置，然后又从前往后比较
            {
                end--;
                len--;
            }
            //从前往后比较
            while(end>=start&&nums[start]!=val)//如果没有等于val的，比较下一个，直到有比等于val的交换位置
                start++;
            if(end>=start){
                int temp = nums[start];
                nums[start] = nums[end];
                nums[end] = temp;
            }
            //一次循环交换一对数值的位置
        }
        // 比较结束，nums即基本有序。左边的值都不等于val，右边的值等于val
        return len;
    }
    public static void main(String[] args){
        int[] nums={1};
        int val=1;
        L27_RemoveElement re=new L27_RemoveElement();
        int len = re.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++) {
            System.out.println(nums[i]);
        }
    }
}