Search in Rotated Sorted Array II - Leetcode

来源：互联网发布：长门号战列舰模型淘宝编辑：程序博客网时间：2024/06/05 15:51

<p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 16px; line-height: 30px;">Follow up for "Search in Rotated Sorted Array":<br style="box-sizing: border-box;" />What if <span style="box-sizing: border-box;">duplicates</span> are allowed?</p><p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 16px; line-height: 30px;">Would this affect the run-time complexity? How and why?</p><p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 16px; line-height: 30px;">Write a function to determine if a given target is in the array.</p>

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public class Solution {    public boolean search(int[] A, int target) {        if(A.length < 1)        return false;        if(A.length < 2)        return A[0] == target;                int low=0, high=A.length-1, mid=(low+high)/2;        while(low <= high){            mid=(low+high)/2;            if(A[mid] == target)            return true;            if(A[mid] > A[low]){                if(A[low]<=target && target<A[mid])                high=mid;                else                low=mid+1;            }else if(A[mid] < A[low]){                if(A[mid]<target && target<=A[high])                low=mid+1;                else                high=mid;            }else            low ++;        }        return false;    }}

思路：不多说了看

Search in Rotated Sorted Array。

tricky的内容就是如果遇到一样的low++知道遇到不同的数字

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