33. Search in Rotated Sorted Array

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.

Cracking Interview 原题,最坏情况下复杂度是O(n)

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

posted on 2015-05-29 08:45  shini  阅读(91)  评论(0编辑  收藏  举报

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