算法导论(第三版)Exercises2.3(归并排序、二分查找、计算集合中是否有和为X的2个元素)

2.3-1:

   3 9 26 38 41 49 52 59

   3 26 41 52   9 38 49 57

3 41   52 26   38 57   9 49

3   41  52  26  38  57  9  49

2.3-2:(归并排序)

void mergeSort(int a[], int l, int r)

{

    int m;

    if(l < r)

    {

        m = (l + r) / 2;

        mergeSort(a, l, m);

        mergeSort(a, m+1, r);

        merge(a, l, r, m);

    }

}



void merge(int a[], int l, int r, int p)

{

    int i, j, k;

    int n1 = p - l + 1;

    int n2 = r - p;

    int lArray[n1], rArray[n2];



    for(i=0; i<n1; i++) lArray[i] = a[l+i];

    for(j=0; j<n2; j++) rArray[j] = a[p+j+1];

    i = 0;

    j = 0;

    for(k=l; k<=r; k++)

    {

        if(i < n1 && (j >= n2 || lArray[i] <= rArray[j]))

        {

            a[k] = lArray[i];

            ++i;

        }

        else if(j < n2 && (i >= n1 || rArray[j] < lArray[i])) 

        {

            a[k] = rArray[j];

            ++j;

        }

    }

}

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2.3-3:

n=2:  Tn=2lg2=2

假设 n=k时等式成立

T_k+1=2(T_k) + 2^k+1=2 * (2^k * k) + 2^k+1=2^k+1(k+1)

2.3-4:

最差情况：

1   n=2;

T_n = T_n-1+n-1;

2.3-5:(二分查找)

int binarySearch(int a[], int v, int n)

{

    int l, r, m;



    l = 0;

    r = n - 1;

    while(l <= r) 

    {

        m = (l + r) / 2;

        if(v < a[m]) r = m - 1;

        else if(v > a[m]) l = m + 1;

        else return m;

    }

    return -1;

}

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最差情况：

2   n=2;

T_n=T(n/2) + 1      n=2^k;

θ(lgn)

2.3-6:

无法降低运行时间

2.3-7:(参考了网友的答案，做了一点优化，希望大家能提更好的改进建议）

算法：求一个N个数的集合是否含有和为X的2个元素(需要支持c99标准)

bool sumX(int a[], int n, int x)

{

    int length, complement[n];

    mergeSort(a, 0, n-1);

    length = deduplication(a, n);

    xComplement(a, complement, length, x);

    return mergeSearch(a, complement, length);

}



bool mergeSearch(int a[], int b[], int n)

{

    int i, j;

    for(i=0, j=n-1; i<n && j>=0 && a[i]!=b[j]; )

        if(a[i] < b[j]) i++;

        else j--;

    return i<n && j>=0;

}



void xComplement(int a[], int aComplement[], int n, int x)

{

    int i;

    for(i=0; i<n; i++) aComplement[i] = x - a[i];

}



int deduplication(int a[], int n)

{

    int i, tmp[n];

    int j = 0;

    for(i=0; i<n; i++)

    {

        if(i > 0 && a[i-1] != a[i]) j++;

        tmp[j] = a[i];

    }

    for(i=0; i<=j; i++) a[i] = tmp[i];

    return j+1;

}



void mergeSort(int a[], int l, int r)

{

    int m;

    if(l < r)

    {

        m = (l + r) / 2;

        mergeSort(a, l, m);

        mergeSort(a, m+1, r);

        merge(a, l, r, m);

    }

}



void merge(int a[], int l, int r, int m)

{

    int max = 1000;

    int i, j, k;

    int n1 = m - l + 1;

    int n2 = r - m;

    int lArray[n1+1], rArray[n2+1];



    for(i=0; i<n1; i++) lArray[i] = a[l+i];

    for(j=0; j<n2; j++) rArray[j] = a[m+j+1];

    lArray[n1] = max;

    rArray[n2] = max;

    i = 0;

    j = 0;

    for(k=l; k<=r; k++)

    {

        if(i < n1 && lArray[i] <= rArray[j])

        {

            a[k] = lArray[i];

            ++i;

        }

        else 

        {

            a[k] = rArray[j];

            ++j;

        }

    }

}

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θ(nlgn)