【C++】【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.

思路

数组其实就是原来的数组分成了两部分，判断target是在前半部分还是后半部分，然后按照二分查找即可。

代码

第一次代码：LeetCode上的代码写的很简洁，但是按照我自己的思路写出来的代码就是分了很多判断分支。

class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0;
int right = nums.size()-1;

while (left < right) {
int mid = (left + right) / 2;
if (target == nums[mid]) {
return mid;
} else if (target == nums[0]) {
return 0;
} else if ((target > nums[0]) && (nums[mid] > nums[0]) && (target < nums[mid])) {
right = mid - 1;
} else if (target > nums[0] && nums[mid] < nums[0]){
right = mid - 1;
} else if ((target > nums[0]) && (nums[mid] > nums[0]) && (target > nums[mid])) {
left = mid + 1;
} else if (target < nums[0] && nums[mid] > nums[0]) {
left = mid + 1;
} else if (target < nums[0] && nums[mid] < nums[0] && target < nums[mid]) {
right = mid - 1;
} else if (target < nums[0] && nums[mid] < nums[0] && target > nums[mid]) {
left = mid + 1;
} else {
left++;
}
}
if (left == right && target == nums[left]) {
return left;
}
return -1;
}
};

第二次代码：把之前的代码稍稍进行了修改，看上去略清楚一些。

class Solution {
public:
int search(vector<int>& nums, int target) {
int left = 0;
int right = nums.size()-1;

while (left < right) {
int mid = (left + right) / 2;
if (target == nums[mid]) {
return mid;
} else if (target == nums[0]) {
return 0;
} else if (
(nums[mid] > target && target > nums[0])
|| (target > nums[0] && nums[0] > nums[mid])
|| (nums[0] > nums[mid] && nums[mid] > target)){
right = mid - 1;
} else if (
(target > nums[mid] && nums[mid] > nums[0])
|| (nums[mid] > nums[0] && nums[0] > target)
|| (nums[0] > target && target > nums[mid]) ) {
left = mid + 1;
} else {
left++;
}
}
if (left == right && target == nums[left]) {
return left;
}
return -1;
}
};