码迷,mamicode.com
首页 > 其他好文 > 详细

Search in Rotated Sorted Array

时间:2018-01-05 22:32:46      阅读:213      评论:0      收藏:0      [点我收藏+]

标签:some   arch   pos   dex   exist   otherwise   com   []   targe   

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

Search in Rotated Sorted Array

标签:some   arch   pos   dex   exist   otherwise   com   []   targe   

原文地址:https://www.cnblogs.com/bingo2-here/p/8207039.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!