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  • Leetcode 27——Remove Element

      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 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:

    Given nums = [3,2,2,3], val = 3,
    
    Your function should return length = 2, with the first two elements of nums being 2.

    分析:in-place就地remove元素,不允许分配另一个数组。可以这么想,允许用另一个数组的话,直接遍历这个数组,满足条件就添加到另一个数组中。但是不让用多一个数组的话,就用自身,因为满足条件的数量肯定是小于等于数组的长度,所以用一个index来计数即可,满足条件的放到index位置上,index++。
    class Solution {
            public int removeElement(int[] nums, int val) {
                int index=0;
                for(int i=0;i<nums.length;i++){
                    if(nums[i]!=val){
                        nums[index]=nums[i];
                        index++;
                    }
                }
                return index;
                
            }
       
    }
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  • 原文地址:https://www.cnblogs.com/GoForMyDream/p/8522704.html
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