摘要:难度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。比如全部是的数列,和除了某位置有个,其余全部是的数列,都是合法的。在这里,测试用例也进行了增加,尽量覆盖各种奇葩情况。
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 case return -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 array System.out.println("----- test non-repeatable shift array -----"); int attemptSize = 100; int randomRange = 999; Random rdm = new Random(); Setts = new TreeSet (); 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 array System.out.println("----- test repeatable shift array -----"); attemptSize = 100; randomRange = 30; shift = rdm.nextInt(attemptSize); System.out.println("size: " + attemptSize + "; shift: " + shift); List jay = new ArrayList (); 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 case return -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; } } }
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