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  • 442.Find All Duplicates in an Array

    Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

    Find all the elements that appear twice in this array.

    Could you do it without extra space and in O(n) runtime?

    Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

    Find all the elements that appear twice in this array.

    Could you do it without extra space and in O(n) runtime?

    Example:

    Input:
    [4,3,2,7,8,2,3,1]
    
    Output:
    [2,3]
    方法一:先进行排序,然后遍历看是否有重复的数
    var findDuplicates = function(nums) {
        var len = nums.length;
        var arr = [];
        nums.sort(function(a,b){
            return a-b;
        })
        for(var i = 0;i < len-1;){
            if(nums[i] == nums[i+1]){
                arr.push(nums[i]);
                i += 2;
            }else{
                i++;
            }
        }
        return arr;
    };

    方法二:对于每个nums[i],将其对应的nums[nums[i] - 1]取相反数,如果其已经是负数了,说明之前存在过,将其加入结果arr中即可

    var findDuplicates = function(nums) {
        var len = nums.length;
        
        var arr = [];
       
        for(var i = 0;i < len;i++){
           var temp = Math.abs(nums[i])-1;
           if(nums[temp] < 0){
               arr.push(temp+1);
           }
           nums[temp] = -nums[temp];
        }
        return arr;
    };
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  • 原文地址:https://www.cnblogs.com/deerfig/p/6675605.html
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