33. Search in Rotated Sorted Array

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.

 

 

public int search(int[] nums, int target) {
    if (nums.length == 0)
        return -1;
    int minIdx = findMinIdx(nums); //先找出最小值的索引
    if (target == nums[minIdx]) 
        return minIdx;
    int m = nums.length;
    int start = (target > nums[m - 1]) ? 0 : minIdx;
    int end = (target > nums[m - 1]) ? minIdx - 1 : m - 1;
    while (start <= end) {
        int mid = start + (end - start) / 2;
        if (nums[mid] == target)
            return mid;
        else if (target > nums[mid]) 
            start = mid + 1;
        else 
            end = mid - 1;
    }
    return -1;
}

public int findMinIdx(int[] array) {
    int low = 0;
    int high = array.length - 1;
    while (low <= high) {
        int mid = low + (high - low) / 2;
        if (array[mid] > array[high])
            low = mid + 1;
        else if (array[mid] == array[high])
            high = high - 1;
        else
            high = mid;
    }
    return low;
}