Follow up for “Find Minimum in Rotated Sorted Array”:
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
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).
Find the minimum element.
The array may contain duplicates.

难度: Hard

就是说一个从小到大排序好的数组循环移位不知多少次，求最小值。数组可以有重复值！
这就比无重复的难一些了。
可以重复会带来不少问题，之前的不重复循环移位的判定条件都要重新思考是否有效。
比如全部是1的数列，和除了某位置有个2，其余全部是1的数列，都是合法的。
上个题目中的移动和终止条件似乎都无效了

相比上题而言，这里有几个针对重复值的关键：

• 如果左游标遇到重复值，则移动到该重复值序列的最右边（moveRight）

• 如果右游标遇到重复值，则移动到该重复值序列的最左边（moveLeft）

还有一个边界情况：如果碰到全1的序列呢？moveLeft会一直进行下去死循环了。所以对于这种情况我们如果循环一圈发现还没有停止moveLeft的意思，就要break掉并且在返回值中进行标记了。

这个算法的时间复杂度相对于上个算法有什么变化呢？平均时间复杂度还是O(log n)，但是出现了最坏的情况，全部重复或者很多重复的情况。最坏肯定是O(n)啦。上一个算法似乎不会出现这种情况吧。

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Random;
import java.util.Set;
import java.util.TreeSet;
public class ShiftFinder2 {public static int findMin(int[] array) {if (array.length == 0) {return 0;}if (array.length == 1) {return array[0];}int len = array.length;int l = 0;int r = len - 1;int cur = (l + r) / 2;while (true) {cur = moveLeft(array, cur);if (cur == -1) {cur = 0;break;}l = moveRight(array, l);r = moveLeft(array, r);if (array[cur] < array[index(cur - 1, len)]) {break;}if (l == r) {cur = l;break;}if (r == (l + 1)) {if (array[l] < array[r]) {cur = l;} else {cur = r;}break;}if (array[cur] < array[l]) {r = cur;cur = (l + r) / 2;continue;}if (array[cur] > array[r]) {l = moveRight(array, cur);cur = (l + r) / 2;continue;}r = cur;cur = (l + r) / 2;}return array[cur];}public static int moveLeft(int[] array, int pos) {int counter = 0;int len = array.length;pos = index(pos, len);while (true) {if (array[pos] != array[index(pos - 1, len)]) {return pos;}if (counter >= len) {// special casereturn -1;}pos = index(pos - 1, len);counter++;}}public static int moveRight(int[] array, int pos) {int len = array.length;pos = index(pos, len);while (true) {if (array[pos] != array[index(pos + 1, len)]) {return pos;}pos = index(pos + 1, len);}}public static int index(int cur, int length) {return (cur % length + length) % length;}public static void main(String[] args) {int[] a = { 7, 8, 11, 12, 13, 14, 19, 22, 1, 2, 4, 5 };int[] b = { 1, 2, 3, 4, 5, 6, 7 };int[] c = { 11, 1, 2, 4, 5, 7, 8 };int[] d = { 1 };int[] e = { 1, 2 };int[] f = { 2, 1 };int[] g = { 3, 1, 2 };int[] h = { 1, 1, 1, 1, 1, 1 };int[] m = { 2, 1, 1, 1, 1, 1, 1 };int[] n = { 2, 2, 1, 1, 1, 1, 1, 1, 2 };int[] o = { 1, 1, 1, 1, 1, 1, 1, 2 };int[] p = { 1, 1, 1, 1, 1, 1, 2, 2 };int[] q = { 5, 5, 1, 3, 3, 3, 3, 3, 3, 4 };int[] r = { 4, 4, 5, 6, 6, 6, 6, 7, 9, 1, 1, 2 };int[] s = { 18, 18, 18, 19, 19, 19, 20, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 23, 23, 23, 23, 23,24, 24, 25, 25, 25, 25, 26, 27, 27, 27, 27, 28, 28, 29, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6,6, 7, 8, 8, 9, 9, 9, 9, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14,14, 15, 15, 16, 17, 17, 17, 17, 17, 17, 18, 18 };int[] t = { 7, 7, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 11, 11, 12, 12, 12, 12, 12, 13, 14, 14, 15, 15, 15, 15,16, 16, 17, 17, 17, 17, 18, 18, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 22, 22, 22,23, 23, 23, 23, 24, 24, 25, 26, 28, 28, 28, 28, 29, 29, 29, 29, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3,3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7 };int[] u = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1 };int[] v = { 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };System.out.println(ShiftFinder2.findMin(a));System.out.println(ShiftFinder2.findMin(b));System.out.println(ShiftFinder2.findMin(c));System.out.println(ShiftFinder2.findMin(d));System.out.println(ShiftFinder2.findMin(e));System.out.println(ShiftFinder2.findMin(f));System.out.println(ShiftFinder2.findMin(g));System.out.println(ShiftFinder2.findMin(h));System.out.println(ShiftFinder2.findMin(m));System.out.println(ShiftFinder2.findMin(n));System.out.println(ShiftFinder2.findMin(o));System.out.println(ShiftFinder2.findMin(p));System.out.println(ShiftFinder2.findMin(q));System.out.println(ShiftFinder2.findMin(r));System.out.println(ShiftFinder2.findMin(s));System.out.println(ShiftFinder2.findMin(t));System.out.println(ShiftFinder2.findMin(u));System.out.println(ShiftFinder2.findMin(v));// gen non-repeatable random shift arraySystem.out.println("----- test non-repeatable shift array -----");int attemptSize = 100;int randomRange = 999;Random rdm = new Random();Set<Integer> ts = new TreeSet<Integer>();for (int i = 0; i < attemptSize; i++) {ts.add(rdm.nextInt(randomRange));}int shift = rdm.nextInt(ts.size());System.out.println("size: " + ts.size() + "; shift: " + shift);Integer[] iay = new Integer[ts.size()];ts.toArray(iay);int[] aa = new int[ts.size()];for (int i = 0; i < ts.size(); i++) {aa[ShiftFinder2.index(i + shift, aa.length)] = iay[i];}for (int i = 0; i < aa.length; i++) {System.out.print(aa[i] + " ");}System.out.println();System.out.println("non-repeatable random minimum find: " + ShiftFinder2.findMin(aa));System.out.println();// gen repeatable random shift arraySystem.out.println("----- test repeatable shift array -----");attemptSize = 100;randomRange = 30;shift = rdm.nextInt(attemptSize);System.out.println("size: " + attemptSize + "; shift: " + shift);List<Integer> jay = new ArrayList<Integer>();for (int i = 0; i < attemptSize; i++) {jay.add(rdm.nextInt(randomRange));}Collections.sort(jay);int[] bb = new int[attemptSize];for (int i = 0; i < attemptSize; i++) {bb[ShiftFinder2.index(i + shift, attemptSize)] = jay.get(i);}for (int i = 0; i < attemptSize; i++) {System.out.print(bb[i] + " ");}System.out.println();System.out.println("repeatable random minimum find: " + ShiftFinder2.findMin(bb));System.out.println();}
}

在这里，测试用例也进行了增加，尽量覆盖各种奇葩情况。后面除了不重复的随机移位数列之外，还有可重复的随机移位序列。

提交的代码:

public class Solution {public int findMin(int[] array) {if (array.length == 0) {return 0;}if (array.length == 1) {return array[0];}int len = array.length;int l = 0;int r = len - 1;int cur = (l + r) / 2;while (true) {cur = moveLeft(array, cur);if (cur == -1) {cur = 0;break;}l = moveRight(array, l);r = moveLeft(array, r);if (array[cur] < array[(cur+len-1)%len]) {break;}if (l == r) {cur = l;break;}if (r == (l + 1)) {if (array[l] < array[r]) {cur = l;} else {cur = r;}break;}if (array[cur] < array[l]) {r = cur;cur = (l + r) / 2;continue;}if (array[cur] > array[r]) {l = moveRight(array, cur);cur = (l + r) / 2;continue;}r = cur;cur = (l + r) / 2;}return array[cur];        }public static int moveLeft(int[] array, int pos) {int counter = 0;int len = array.length;while (true) {if (array[pos] != array[(pos+len-1)%len]) {return pos;}if (counter >= len) {// special casereturn -1;}pos = (pos+len-1)%len;counter++;}}public static int moveRight(int[] array, int pos) {int len = array.length;while (true) {if (array[pos] != array[(pos+1)%len]) {return pos;}pos = (pos+1)%len;}}
}