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 7might become4 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.

Analysis

每次按mid去break down, 总是能把array分成两个部分, 而其中一个部分是sorted的.

Solution

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