【算法习题】数组中任意2个（3个）数的和为sum的组合

题1、给定一个int数组，一个数sum，求数组中和为sum的任意2个数的组合

@Test
public void test_find2() {
    int[] arr = { -1, 0, 2, 3, 4, 7, 8, 9, 10 };
    int sum = 9;
    Arrays.sort(arr);
    List<TwoTuple<Integer, Integer>> result = new ArrayList<>();

    int i = 0;
    int j = arr.length - 1;
    while (i < j) {
        if (arr[i] + arr[j] == sum) {
            result.add(new TwoTuple<Integer, Integer>(arr[i], arr[j]));
            i++;
            j--;
        } else if (arr[i] + arr[j] < sum) {
            i++;
        } else { // > sum
            j--;
        }
    }
    System.out.println(result);
} // out: [-1:10, 0:9, 2:7]

题2、给定一个int数组，一个数sum，求数组中和为sum的任意3个数的组合 

@Test
public void test_find3() {
    int[] arr = { -1, 0, 2, 3, 4, 7, 8, 9, 10 };
    int sum = 9;
    Arrays.sort(arr);
    Set<ThreeTuple<Integer, Integer, Integer>> result = new LinkedHashSet<>();

    for (int i = 0; i < arr.length; i++) {
        int temp = arr[i];
        swap(arr, i, 0);
        int start = 1;
        int end = arr.length - 1;
        while (start < end) {
            if (arr[start] + arr[end] == sum - temp) {
                int[] threeNum = {temp, arr[start], arr[end]};
                Arrays.sort(threeNum);
                result.add(new ThreeTuple<>(threeNum[0], threeNum[1], threeNum[2]));
                start++;
                end--;
            } else if (arr[start] + arr[end] > sum - temp) {
                end--;
            } else {
                start++;
            }
        }
        swap(arr, i, 0); // 还原
    }
    System.out.println(result);
} // out: [-1:0:10, -1:2:8, -1:3:7, 0:2:7, 2:3:4]

private void swap(int[] a, int i, int j) {
    int temp = a[i];
    a[i] = a[j];
    a[j] = temp;
}

上面两题用到的元组类： 

class TwoTuple<A, B> {
    public final A first;
    public final B second;

    public TwoTuple(A a, B b) {
        this.first = a;
        this.second = b;
    }
    
    @Override public String toString() {
        return first + ":" + second;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((first == null) ? 0 : first.hashCode());
        result = prime * result + ((second == null) ? 0 : second.hashCode());
        return result;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        TwoTuple<?, ?> other = (TwoTuple<?, ?>) obj;
        if (first == null) {
            if (other.first != null)
                return false;
        } else if (!first.equals(other.first))
            return false;
        if (second == null) {
            if (other.second != null)
                return false;
        } else if (!second.equals(other.second))
            return false;
        return true;
    }
}

class ThreeTuple<A, B, C> extends TwoTuple<A, B> {
    public final C third;

    public ThreeTuple(A a, B b, C c) {
        super(a, b);
        this.third = c;
    }
    
    @Override public String toString() {
        return super.toString() + ":" + third;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = super.hashCode();
        result = prime * result + ((third == null) ? 0 : third.hashCode());
        return result;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (!super.equals(obj))
            return false;
        if (getClass() != obj.getClass())
            return false;
        ThreeTuple<?, ?, ?> other = (ThreeTuple<?, ?, ?>) obj;
        if (third == null) {
            if (other.third != null)
                return false;
        } else if (!third.equals(other.third))
            return false;
        return true;
    }
}

题3、给定一个int正整数数组，一个数sum，求数组中和为sum的k个数的组合有多少种（k任意）。 

另开一博客讨论这个问题。

指路：【算法习题】正整数数组中和为sum的任意个数的组合数——待完成