LeetCode Remove Element

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原题链接在这里：https://leetcode.com/problems/remove-element/

题目：

Given an array and a value, 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 in place with constant memory.

The order of elements can be changed. It doesn‘t matter what you leave beyond the new length.

Example:
Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

题解：

类似Move Zeroes.

遇到不等于val的值就放到nums[count]上，同事count++.

Time Complexity: O(nums.length). Space: O(1).

AC Java:

 1 public class Solution {
 2     public int removeElement(int[] nums, int val) {
 3         if(nums == null || nums.length == 0){
 4             return 0;
 5         }
 6         int count = 0;
 7         for(int i = 0; i<nums.length; i++){
 8             if(nums[i]!= val){
 9                 nums[count++] = nums[i];
10             }
11         }
12         return count;
13     }
14 }

LeetCode Remove Element

标签：array   www   ons   new   complex   first   space   log   bsp   

原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/6247239.html