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TwoSum:两数相加得0

在一个不重复的数组中,统计有多少组两个元素相加得0。

这里使用三种方式实现,并统计他们各自花费的时间:

import java.util.Arrays;import java.util.HashMap;import java.util.Random;public class TwoSum {	private static int N = 100000;	private static int[] a = new int[N];	private static Random random = new Random();	private static HashMap<Integer, Boolean> m = new HashMap<Integer, Boolean>();		private int binarySearch(int[] a, int num) {		int low = 0, mid = 0, high = a.length - 1;		while (low <= high) {			mid = low + (high - low) / 2;			if (a[mid] > num) {				high = mid - 1;			} else if (a[mid] < num) {				low = mid + 1;			} else {				return mid;			}		}				return -1;	}		public int solution1() {		int count = 0;		for (int i = 0; i < N; i++) {			for (int j = i + 1; j < N; j++) {				if (a[i] + a[j] == 0) {					count++;				}			}		}		return count;	}		public int solution2() {		int count = 0;		Arrays.sort(a);		for (int i = 0; i < N; i++) {			if (binarySearch(a, -a[i]) > i) {				count++;			}		}		return count;	}		public int solution3() {		int count = 0;		for (int i = 0; i < N; i++) {			if (m.containsKey(-a[i])) {				count++;			}		}		return count / 2;	}		public static int uniform(int N) {        if (N <= 0) throw new IllegalArgumentException("Parameter N must be positive");        return random.nextInt(N);    }		public int uniform(int a, int b) {        if (b <= a) throw new IllegalArgumentException("Invalid range");        if ((long) b - a >= Integer.MAX_VALUE) throw new IllegalArgumentException("Invalid range");        return a + uniform(b - a);    }		public static void main(String[] args) {		TwoSum ts = new TwoSum();		int count = 0;		int rand = 0;		int i = 0;		boolean flag = true;				while (i < a.length) {			rand = ts.uniform(-10000000, 10000000);			flag = true;			for (int j = 0; j < a.length; j++) {				if (a[j] == rand) {					flag = false;					break;				}			}			if (flag) {				a[i] = rand;							i++;				//System.out.println(rand + " " + flag);			}		}				for (int j = 0; j < N; j++) {			m.put(a[j], true);		}				Stopwatch timer1 = new Stopwatch();		count = ts.solution1();		double time = timer1.elapsedTime();		System.out.println("count1: " + count + " cost1: " + time);				count = 0;		time = 0;		Stopwatch timer2 = new Stopwatch();		count = ts.solution2();		time = timer2.elapsedTime();		System.out.println("count2: " + count + " cost2: " + time);				count = 0;		time = 0;		Stopwatch timer3 = new Stopwatch();		count = ts.solution3();		time = timer3.elapsedTime();		System.out.println("count3: " + count + " cost3: " + time);	}}

solution1:直接使用一个两层for循环,所以最终的时间复杂度为O(N^2);

solution2:先将数组排序,然后通过二分查找来搜索数组中每个元素的相反数,此时的时间复杂度为O(N*log(N));

solution3:和方法一类似,只不过是通过哈希表中查找元素的复杂度为O(1)来优化搜索时间,此时时间复杂度为O(N);

当N=1000时count1: 0 cost1: 0.004count2: 0 cost2: 0.002count3: 0 cost3: 0.001当N=10000时count1: 5 cost1: 0.026count2: 5 cost2: 0.009count3: 5 cost3: 0.005当N=100000时count1: 238 cost1: 2.296count2: 238 cost2: 0.077count3: 238 cost3: 0.012

这里因为产生不重复的随机数使用的是最简单的两层循环,所以当数组的length太大时,比如1000000时,在产生随机数的过程中会消耗很长时间,这一部分可以使用其他更优的方法实现。

TwoSum:两数相加得0