Java 旋转数组查找旋转点和任意元素(元素可重复)

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x

private 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));
}
}