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  • [leetcode]Search in Rotated Sorted Array

    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.

    二分查找的变形

    算法思路:

    a[mid] == target return mid;

    a[mid] < target 分为两种情况

    1. a[mid]和target在同半边,begin = mid + 1

    2. a[mid]和target在不同的半边,则a[mid]肯定在后面,target在前半边,因此往前找end = mid - 1;

    a[mid] > target同理

    1. a[mid]和target在同半边,end = mid - 1;

    2. a[mid]和target在不同的半边,则a[mid]肯定在前面,target在后半边,因此往前找begin = mid + 1;

    代码如下:

     1 public class Solution {
     2     public int search(int[] a, int target) {
     3         if(a == null || a.length == 0) return -1;
     4         int length = a.length;
     5         int begin = 0, end = length - 1;
     6         while(begin <= end){
     7             int mid = (begin + end) >> 1;
     8             if(a[mid] == target){
     9                 return mid;
    10             }else if(a[mid] > target){
    11                 if(target < a[begin] && a[mid] >= a[begin]){//这里如果写成a[mid]<=a[end]就不行了,因为a[mid]<=a[end]不能保证在前半段
    12                     begin = mid + 1;
    13                 }else{
    14                     end = mid - 1;
    15                 }
    16             }else{
    17                 if(a[mid] < a[begin] && target >= a[begin]){
    18                     end = mid - 1;
    19                 }else{
    20                     begin = mid + 1;
    21                 }
    22             }
    23         }
    24         return -1;
    25     }
    26 }
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  • 原文地址:https://www.cnblogs.com/huntfor/p/3901303.html
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