二分查找

//二分查找，binary search
//给定一个整数X和整数A0，A1，A2...AN-1,后者已排序（从小到大），求使得Ai=X的下标i，
//如果X不在数据中，则返回-1.
//时间复杂度O(logN),空间复杂度O（1）
int BinarySearch(int nums[], int n, int x){
int first, last, mid;
first = 0;
last = n - 1;

//我自己写的，但代码不够简洁
/*while (first != last){
mid = (first + last) / 2;
if (nums[first] <= x && x < nums[mid]){
last = mid;
}
else{
first = mid + 1;
}
}
if (nums[mid] == x){
return mid;
}
else{
return -1;
}*/

//参照《数据结构与算法分析》，比较简洁明了
while (first <= last){
mid = (first + last) / 2;
if (x < nums[mid]){
last = mid;
}
else if (x>nums[mid]){
first = mid + 1;
}
else{
return mid;
}
}
return -1;
}

//Search in Rotated Sorted Array

/*描述
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
*/
//从小到大排序，
/*分析
二分查找，难度主要在于左右边界的确定。
*/
//时间复杂度O(logN),空间复杂度O(1).
int Search(int nums[], int n, int x){
int first, last, mid;
first = 0;
last = n - 1;
while (first <= last){
mid = (first + last) / 2;
if (x == nums[mid]){
return mid;
}

if (nums[first] <= nums[mid]){
//first部分没有rotate(分析错误)
//first 到 Mid 递增，
if (nums[first]<=x && x < nums[mid]){
last = mid;
}
else{
first = mid + 1;
}
}
else{
//nums[first]>nums[mid]
//first是从后面rotate来的，mid到last递增。
if (nums[mid] < x && x <= nums[last - 1]){
first = mid + 1;
}
else{
last = mid;
}
}
}
return -1;
}

/*Search in Rotated Sorted Array II*/
/*
描述
Follow up for ”Search in Rotated Sorted Array”: What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
*/
//时间复杂度O(n),空间复杂度O(1).
bool Search2(int nums[], int n, int x){
int first, last, mid;
first = 0;
last = n - 1;
while (first <= last){
mid = (first + last) / 2;
if (x == nums[mid]){
return true;
}
else{
if (nums[first] < nums[mid]){
//first to mid is 递增
if (nums[first] <= x && x < nums[mid]){
last = mid;
}
else{
first = mid + 1;
}
}
else if (nums[first]>nums[mid]){
//mid to last is 递增
if (nums[mid] < x && x <= nums[last]){
first = mid + 1;
}
else{
last = mid;
}
}
else{
//nums[first]=nums[mid]
//first 有重复，skip duplicate one
first++;
}
}
}
return false;
}