【二分】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.

方法一：先找出旋转数组的最小节点，再进行分段二分查找

public class Solution {        public int findPivot(int [] a){//找出旋转数组的最小值下标        int l = 0;        int r = a.length-1;        int pivot = l;                while(l<r && a[l] >= a[r]){//注意条件        if(r - l == 1){        pivot = r;        break;        }            int mid = (l+r)/2;            if(a[mid] <= a[l]) r = mid;//和二分查找的变化有区别            else l = mid;        }        return pivot;    }        public int binaySearch(int [] a, int target, int l, int r){//真二分查找        while(l <= r){            int mid = (l+r)/2;            if(a[mid] < target) l = mid+1;            else if(a[mid] > target) r = mid-1;            else return mid;        }        return -1;    }        public int search(int[] A, int target) {        if(A.length == 0) return -1;        int pivot = findPivot(A);        //System.out.println(pivot);        int res1 = binaySearch(A, target, 0, pivot-1);        if(res1 != -1) return res1;        else return binaySearch(A, target, pivot, A.length-1);    }}

方法二：递归查找

public int search(int[] A, int target) {  return find(A, 0, A.length - 1, target);}private int find(int[] arr, int p, int q, int x) {  if (p > q)    return -1;  int mid = (p + q) / 2;  if (arr[mid] == x)    return mid;  if (arr[p] < arr[mid]) { // left is ordered, right is broken    if (arr[p] <= x && x <= arr[mid]) // x is in the left ordered part      return find(arr, p, mid - 1, x);    else // x doesn't belong to left ordered part      return find(arr, mid + 1, q, x);  } else if (arr[p] > arr[mid]) {// left is broken, so right is ordered    if (arr[mid] <= x && x <= arr[q])      return find(arr, mid + 1, q, x);    else      return find(arr, p, mid - 1, x);  } else { // a[p] == a[mid], directly search right      return find(arr, mid + 1, q, x);  }}