算法分析与设计——LeetCode:33. Search in Rotated Sorted Array

题目

Suppose an array sorted in ascending order 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.

class Solution {
public:
int search(vector<int>& nums, int target) {
}
};

思路

使用二分查找，有三种情况：1.mid>begin>end；2.mid<end<begin；3.begin<mid<end，即正常排序的数列。 根据不同情况选择下一次查找的begin和end。

代码

class Solution {
public:
int search(vector<int>& nums, int target) {
int end = nums.size();
if (end == 0) {
return -1;
}
int head = 0;
int index = end/2;

while ((end-head) != 0 && nums[index] != target) {
if (nums[head] < nums[end-1]) {
while ((end-head) != 0 && nums[index] != target) {
if (target > nums[index]) {
head = index+1;
} else {
end = index;
}
index = (end-head)/2+head;
}
break;
}

if(nums[index] > nums[head]) {
if (target > nums[index]) {
head = index +1;
} else {
if (target >= nums[head]) {
end = index;
} else {
head = index+1;
}
}
} else {
if (target > nums[index]) {
if (target >= nums[head]) {
end = index;
} else {
head = index+1;
}
} else {
end = index;
}
}
index = (end-head)/2+head;
}
if (target != nums[index]) {
return -1;
} else {
return index;
}

}
};