leetcode 33. Search in Rotated Sorted Array __java

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.

思路：

i为左端点 j为右端点 (i+j)/2 = mid为中间点 起始点位置为x

1 左端点为起始点 x = i



2 i < x < mid



3 x = mid



4 mid < x < j



5 x = j



虽然有点浪费时间但是画出来之后边界看的还是比较清晰

网上一个博客用3个条件 将23分为一种 45分为一种：

1.vi < vj 就是第一种

2.v mid < vi 且 v mid < vj 是第二三种

3 v mid > vi 且 v mid > vj 的是第四五种

差不多可以 了 接下来就是target 与 vi vmid vj 比 以确定在哪个区间了 递归二分查找

与普通有序数组的二分查找的不同就是多了几个条件来区分起始点情况

public int search(int[] nums, int target) {
int i = 0;
int j = nums.length - 1;
int mid = (i+j) / 2;
while(i <= j){
mid = (i + j) / 2;
if(target == nums[mid]) return mid;
if(nums[i] < nums[j]){
if(target < nums[mid]) j = mid - 1;
else i = mid + 1;
}
else if(nums[mid] < nums[j]){
if(target > nums[mid] && target <= nums[j]) i = mid + 1;
else j = mid - 1;
}
else{
if(target < nums[mid] && target >= nums[i]) j = mid - 1;
else i = mid + 1;
}
}
return -1;
}