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  • [leetcode33Search in Rotated Sorted Array]在排序旋转后序列中找目标值

    直接上代码

    /**
     * Created by lvhao on 2017/6/30.
     * 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.
     思路是目标值肯定在一个升序子序列中(可能子序列很短,但是肯定存在),目标就是找到这个子序列然后用二分法找到值
     找目标序列的时候也是用二分法,判断的依据是mid值比fin值大的话说明左边是升序,小的话,右边是升序,然后在利用升序序列的的
     首尾值判断目标值是否在里边,在里边就用二分法找值,不在里边就改变fin或者sta值,继续循环找,注意二分法的时候while的判断
     条件是sta<= fin,如果没有=的话,两个数的情况判断不出来。
     */
    public class Q33SearchRotatedArray {
        public static void main(String[] args) {
            int[] nums = new int[]{3,1};
            int target = 1;
            int res = search(nums,target);
            System.out.println(res);
    
        }
        public static  int search(int[] nums, int target) {
            int len = nums.length;
            //判断特殊情况
            if(len == 1 && nums[0] == target)
            {
                return 0;
            }
            int sta = 0,fin = len-1,mid,index = -1;
            //注意等号
            while(sta <= fin)
            {
                mid = (sta + fin)/2;
                if(nums[mid] < nums[fin])
                {
                    if(target <= nums[fin] && target >= nums[mid])
                    {
                        index = binarySearch(nums,mid,fin,target);
                        //找到值之后立即break
                        break;
                    }
    
                    else
                        fin = mid -1;
                }
                else
                {
                    if(target <= nums[mid] && target >= nums[sta])
                    {
                        index = binarySearch(nums,sta,mid,target);
                        break;
                    }
    
                    else
                        sta = mid + 1;
                }
            }
            return index;
        }
        public static int binarySearch(int[] nums,int sta,int fin,int target)
        {
            int mid,index = -1;
            //注意等号
            while (sta <= fin)
            {
                mid = (sta + fin)/2;
                if(target == nums[mid])
                {
                    index = mid;
                    break;
                }
                else if (target < nums[mid])
                    fin = mid-1;
                else
                    sta = mid+1;
            }
            return index;
        }
    }
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  • 原文地址:https://www.cnblogs.com/stAr-1/p/7099643.html
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