Search in Rotated Sorted Array II 旋转数组（有重复元素）里找数@LeetCode

与之前的题目代码一样。。

是上一题的followup http://blog.csdn.net/fightforyourdream/article/details/16854537

package Level5;/** * Search in Rotated Sorted Array II *  * Follow up for "Search in Rotated Sorted Array": What if duplicates are * allowed? *  * Would this affect the run-time complexity? How and why? *  * Write a function to determine if a given target is in the array. *  */public class S81 {public static void main(String[] args) {}public static boolean search(int[] A, int target) {return rec(A, target, 0, A.length-1) != -1;}// 递归查找public static int rec(int[] A, int target, int low, int high) {if (low > high) { // 没找到的情况return -1;}int mid = low + (high - low) / 2;if (A[mid] == target) { // 找到了return mid;}int res = rec(A, target, low, mid - 1); // 在左侧查找if (res == -1) { // 如果左侧没找到，继续在右侧查找res = rec(A, target, mid + 1, high);}return res;}}

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

// it ends up the same as sequential search// We used linear search for this question just to indicate that the // time complexity of this question is O(n) regardless of binary search is applied or not.public class Solution {    public boolean search(int[] A, int target) {        for (int i = 0; i < A.length; i ++) {            if (A[i] == target) {                return true;            }        }        return false;    }}