zoukankan      html  css  js  c++  java
  • LeetCode_27. Remove Element

    27. Remove Element

    Easy

    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]);
    }
    package leetcode.easy;
    
    public class RemoveElement {
    	@org.junit.Test
    	public void test() {
    		int[] nums11 = { 3, 2, 2, 3 };
    		int[] nums12 = { 0, 1, 2, 2, 3, 0, 4, 2 };
    		int[] nums21 = { 3, 2, 2, 3 };
    		int[] nums22 = { 0, 1, 2, 2, 3, 0, 4, 2 };
    		int val1 = 3;
    		int val2 = 2;
    		System.out.println(removeElement1(nums11, val1));
    		System.out.println(removeElement1(nums12, val2));
    		System.out.println(removeElement2(nums21, val1));
    		System.out.println(removeElement2(nums22, val2));
    	}
    
    	public int removeElement1(int[] nums, int val) {
    		int i = 0;
    		for (int j = 0; j < nums.length; j++) {
    			if (nums[j] != val) {
    				nums[i] = nums[j];
    				i++;
    			}
    		}
    		return i;
    	}
    
    	public int removeElement2(int[] nums, int val) {
    		int i = 0;
    		int n = nums.length;
    		while (i < n) {
    			if (nums[i] == val) {
    				nums[i] = nums[n - 1];
    				// reduce array size by one
    				n--;
    			} else {
    				i++;
    			}
    		}
    		return n;
    	}
    }
    
  • 相关阅读:
    frida枚举当前加载的模块以及模块中方法
    python request请求时候json严格校验怎么去除空格
    firda-so静态注册
    LeetCode 724. 寻找数组的中心索引
    LeetCode 679. 24点游戏
    LeetCode 845. 数组中的最长山脉
    并查集各种情况下的时间复杂度
    LeetCode 547. 省份数量
    LeetCode 5. 最长回文子串
    LeetCode 103. 二叉树的锯齿形层序遍历
  • 原文地址:https://www.cnblogs.com/denggelin/p/11550731.html
Copyright © 2011-2022 走看看