LeetCode TwoSum(两和)

LeetCode 每日一题

****LeetCode 每日一题:
给定一个整数数组，返回两个数字的索引，使它们相加到一个特定的目标。
您可以假设每个输入都只有一个解决方案，而您可能不会使用相同的元素两次。
列:给定nums = [2，7，11，15]，target = 9，
因为nums [ 0 ] + nums [ 1 ] = 2 + 7 = 9，
返回[ 0，1 ]。**

public static void main(String[] args) {        //int[] num = new int[] {150,24,79,50,88,345,3};        //int[] num = new int[] {3,3};        //int[] num = new int[]{3,2,4};        //int[] num = new int[]{0,1,2,0};        int[] num = new int[] {230,863,916,585,981,404,316,785,88,12,70,435,384,778,887,                755,740,337,86,92,325,422,815,650,920,125,277,336,221,847,168,23,677,61,                400,136,874,363,394,199,863,997,794,587,124,321,212,957,764,173,314,422,                927,783,930,282,306,506,44,926,691,568,68,730,933,737,531,180,414,751,28,                546,60,371,493,370,527,387,43,541,13,457,328,227,652,365,430,803,59,858,                538,427,583,368,375,173,809,896,370,789};        int target = 666;        if(num.length < 3) {            if(num[0] + num[1] == target) {                System.out.print("0 1");            }        }else{            int[] twoSum = twoSum(num,target);            for (int i = 0; i < twoSum.length   ; i++) {                if(twoSum[0] == twoSum[1]) {                    throw  new  IllegalArgumentException ("No two sum solution");                 }                System.out.print(twoSum[i]+" ");            }        }    }    public static int[] twoSum(int num[],int target) {        int temp  = num.length;        if(num.length > 5) {            temp = num.length - 1 ;        }        int q = 0;        int e = 0;        for (int i = 0; i < temp ; i++) {                for (int j = i+1; j < temp; j++) {                    //先判断是否等于这个值                    if(num[i] + num[j] == target ) {                        q = num[i];                        e = num[j];                    }                }                //找到第一个 继续找下一个                if(num[i] == q) {                      q = i;                     continue;                }                //找到第二个直接中断循环                if(num[i] == e) {                    e = i;                    break;                }        }        return new int[] {q,e};    }

大家有没有更好的思路,让程序运行更快一点呢？