public class Test {// 已知数组A是由有序数组B(数组B未知)向右移动n(0<=n<=B.length)位得到// 例如 A = {4, 5, 1, 2, 3} 由 B = {1, 2, 3, 4, 5} 向右移动2位得到// 求在数组A中搜索元素xprivate static int search(int[] a, int h, int t, int x) {if (h > t) return -1;int result = -1;int mid = (h + t) / 2;if (a[mid] == x) return mid;if (a[h] == a[t] && a[h] == a[mid]) {// 这种情况无法判断哪边是递增，那边是旋转数组，如下// a = {5, X, 5, X, 5}// a = {5, 5, 5, 6, 5}// a = {5, 6, 5, 5, 5}result = search(a, h, mid - 1, x);if (result == -1) result = search(a, mid + 1, t, x);} else if (a[h] < a[t]) {result = binarySearch(a, h, t, x);} else {    // a[h] > a[t] || ( a[h] == a[t] && a[h] != a[mid] )if (a[mid] >= a[h]) {// a[h]~a[mid]为递增，a[mid+1]~a[t]为旋转数组result = x >= a[h] ? binarySearch(a, h, mid - 1, x) : search(a, mid + 1, t, x);} else {// a[h]~a[mid]为旋转数组，a[mid+1]~a[t]为递增result = x >= a[mid + 1] ? binarySearch(a, mid + 1, t, x) : search(a, h, mid - 1, x);}}return result;}// 第二种方法，先找出数组移位的位数，然后判断应该在哪一段进行二分查找private static int getMoveStep(int[] a, int h, int t) {if (h >= t) return 0;int low = h;int high = t;int mid = (low + high) / 2;int step = 0;if (a[low] == a[high]) {// 这是一种无法判断哪边是递增序列的情况step = getMoveStep(a, h, mid - 1);if (step == 0) step = getMoveStep(a, mid + 1, t);} else {while (a[low] > a[high]) {mid = (low + high) / 2;if (a[mid] >= a[low]) {// 递增序列在mid左边step = low = mid + 1;} else {// 递增序列在mid右边high = mid - 1;}}}return step;}// 利用y移位次数查找private static int searchByMove(int[] a, int x) {int result;int step = getMoveStep(a, 0, a.length - 1);if (step != 0) {result = x > a[0] ? binarySearch(a, 0, step - 1, x) : binarySearch(a, step, a.length - 1, x);} else {result = binarySearch(a, 0, a.length - 1, x);}return result;}private static int binarySearch(int[] a, int h, int t, int x) {int low = h;int high = t;int mid;while (low <= high) {mid = (low + high) / 2;if (a[mid] == x) {return mid;} else if (a[mid] < x) {low = mid + 1;} else {high = mid - 1;}}return -1;}public static void main(String[] args) {int[] a = {4, 5, 1, 2, 3};int[] b = {5, 5, 5, 7, 5};int[] c = {5, 3, 5, 5, 5};int[] d = {5, 5, 3, 4, 5, 5, 5};int[] e = {5, 5, 5, 5, 5};int[] f = {1, 2, 3, 4, 5};System.out.println(search(a, 0, a.length - 1, 1));System.out.println(search(b, 0, b.length - 1, 7));System.out.println(search(c, 0, c.length - 1, 3));System.out.println(search(d, 0, d.length - 1, 3));System.out.println(search(e, 0, e.length - 1, 3));System.out.println(search(f, 0, f.length - 1, 2));System.out.println();System.out.println(searchByMove(a, 1));System.out.println(searchByMove(b, 7));System.out.println(searchByMove(c, 3));System.out.println(searchByMove(d, 3));System.out.println(searchByMove(e, 3));System.out.println(searchByMove(f, 2));}
}