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二分查找及其变形

程序员文章站 2022-07-12 09:35:42
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最基本的二分查找模版:

在有序数组A中查找key,如果找到,返回位置索引,否则,返回-1;

int BinarySearch(int A[], int n, int key)
{
    int left = 0, right = n - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (A[mid] < key) {
            left = mid + 1;
        } else if (A[mid] > key) {
            right = mid - 1;
        } else {// A[mid] == key
            return mid;
        }
    }
    return -1;
}

变种1:如果A有多个key元素,返回最大的,否则,返回-1

int BinarySearch(int A[], int n, int key)
{
    int left = 0, right = n - 1;
    int result = -1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (A[mid] < key) {
            left = mid + 1;
        } else if (A[mid] > key) {
            right = mid - 1;
        } else {// A[mid] == key
            result = mid;
            left = mid + 1;//继续查找右半边
        }
    }
    return result;
}

变种2:如果A中有多个key元素,返回最小的,否则,返回-1

int BinarySearch(int A[], int n, int key)
{
    int left = 0, right = n - 1;
    int result = -1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (A[mid] < key) {
            left = mid + 1;
        } else if (A[mid] > key) {
            right = mid - 1;
        } else {// A[mid] == key
            result = mid;
            right = mid - 1;// 继续查找左半边
        }
    }
    return result;
}

 变种3:在有序数组A中,查找出最小的比key大的元素索引。

int BinarySearch(int A[], int n, int key)
{
   if (A[n-1] < key)
        return -1;
    int left = 0, right = n - 1;
    while (left != right) {
        int mid = left + (right - left) / 2;
        if (A[mid] <= key) {
            left = mid + 1;
        } else {
            right = mid;
        } 
    }
    return left;
}

变种4:在有序数组A中,查找出最大的比k小的元素索引。

int BinarySearch(int A[], int n, int key)
{
    if (A[0] > key)
        return -1;
    int left = 0, right = n - 1;
    while (left < right-1) {
        int mid = left + (right - left) / 2;
        if (A[mid] >= key) {
            right = mid - 1;
        } else {
            left = mid;
        } 
    }
    if (A[right] < key) {
        return right;
    } else {
        return left;
    }
}