二分查找及其变种简单易懂的模版

鉴于最近在网上看到的二分查找算法非常复杂，细节太多，不容易理解，下面给出几个实现简洁，又容易理解的代码模版。

首先，让我们记住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;
    }
}