有两个有序整数集合a和b，写一个函数找出它们的交集？

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方法一:

　private static Set<Integer> setMethod(int[] a,int[] b){         Set<Integer> set = new HashSet<Integer>();         Set<Integer> set2 = new HashSet<Integer>();         for(int i=0; i<a.length; i++) {             set.add(a[i]);         }         for(int j=0; j<b.length; j++) {             if(!set.add(b[j]))                set2.add(b[j]);         }         return set2;     }

方法二:

private static Set<Integer> forMethod(int[] a,int[] b){        Set<Integer> set=new HashSet<Integer>();        int i=0,j=0;        while(i<a.length && j<b.length){            if(a[i]<b[j])                i++;            else if(a[i]>b[j])                j++;            else{                set.add(a[i]);                i++;                j++;            }        }        return set;    }

方法三:

private static int[] intersect(int[] a, int[] b) {        if (a[0] > b[b.length - 1] || b[0] > a[a.length - 1]) {            return new int[0];        }        int[] intersection = new int[Math.max(a.length, b.length)];        int offset = 0;        for (int i = 0, s = i; i < a.length && s < b.length; i++) {            while (a[i] > b[s]) {                s++;            }            if (a[i] == b[s]) {                intersection[offset++] = b[s++];            }            while (i < (a.length - 1) && a[i] == a[i + 1]) {                i++;            }        }        if (intersection.length == offset) {            return intersection;        }        int[] duplicate = new int[offset];        System.arraycopy(intersection, 0, duplicate, 0, offset);        return duplicate;    }

三种性能对比测试:

public class NumberCrossTest {    public static void main(String[] args) {        int[] a1 = new int[100000];        for (int i = 0; i < a1.length; i++) {            a1[i] = i + 10;        }        int[] a2 = new int[200000];        for (int i = 0; i < a2.length; i++) {            a2[i] = i + 20;        }        long begin = System.currentTimeMillis();        Set<Integer> set1 = setMethod(a1, a2);        long end = System.currentTimeMillis();        System.out.println(end - begin);// 359        begin = System.currentTimeMillis();        Set<Integer> set2 = forMethod(a1, a2);        end = System.currentTimeMillis();        System.out.println(end - begin);// 160        begin = System.currentTimeMillis();        int[] c = intersect(a1, a2);        end = System.currentTimeMillis();        System.out.println(end - begin);// 10        // 测试两种方法的结果是否相等        System.out.println(set1.equals(set2));// true        Set<Integer> set3 = new HashSet<Integer>();        for (int i = 0; i < c.length; i++) {            set3.add(c[i]);        }        System.out.println(set1.equals(set3));// true    }    private static Set<Integer> setMethod(int[] a, int[] b) {        Set<Integer> set = new HashSet<Integer>();        Set<Integer> set2 = new HashSet<Integer>();        for (int i = 0; i < a.length; i++) {            set.add(a[i]);        }        for (int j = 0; j < b.length; j++) {            if (!set.add(b[j]))                set2.add(b[j]);        }        return set2;    }    private static Set<Integer> forMethod(int[] a, int[] b) {        Set<Integer> set = new HashSet<Integer>();        int i = 0, j = 0;        while (i < a.length && j < b.length) {            if (a[i] < b[j])                i++;            else if (a[i] > b[j])                j++;            else {                set.add(a[i]);                i++;                j++;            }        }        return set;    }    private static int[] intersect(int[] a, int[] b) {        if (a[0] > b[b.length - 1] || b[0] > a[a.length - 1]) {            return new int[0];        }        int[] intersection = new int[Math.max(a.length, b.length)];        int offset = 0;        for (int i = 0, s = i; i < a.length && s < b.length; i++) {            while (a[i] > b[s]) {                s++;            }            if (a[i] == b[s]) {                intersection[offset++] = b[s++];            }            while (i < (a.length - 1) && a[i] == a[i + 1]) {                i++;            }        }        if (intersection.length == offset) {            return intersection;        }        int[] duplicate = new int[offset];        System.arraycopy(intersection, 0, duplicate, 0, offset);        return duplicate;    }}

结果对比:

方法一用时：359 毫秒

方法二用时：160 毫秒

方法三用时：10 毫秒

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