标签：break int nbsp hint without sea sequence ase rgs
题目要求：
An array is bitonic if it is comprised of an increasing sequence of integers followed immediately by a decreasing sequence of integers. Write a program that, given a bitonic array of n distinct integer values, determines whether a given integer is in the array.
分析
bitonic array是一个内部先递增再递减的队列，实现3lgn很简单，先找出最大值用掉lgn，再从左侧有序数组中查找用掉lgn，未找到则再从右侧有序数组中查找用掉lgn，最坏情况下复杂度就是3lgn。详见代码中findNormal方法。
但是2lgn的做法我一直没想出来合适的，作业的hint提示“Signing bonus. Do it without finding the maximum integer.”意味着去掉查找最大值可以减少复杂度。我一时也没想到好的实现方法，在网上看到一篇别人转载的文章后恍然大悟(http://blog.csdn.net/fiveyears/article/details/11263381)。参照其思路，自己实现findFast方法。
1 public class SearchBitonicArray { 2 private static int seachLeft(int key,int[] a, int lo, int hi){ 3 int mid = 0; 4 while(lo <= hi){//遍历左侧增长序列 5 mid = lo + (hi-lo)/2; 6 if(key > a[mid]) lo = mid +1; 7 else if(key < a[mid]) hi = mid -1; 8 else return mid; 9 } 10 return -1; 11 } 12 private static int searchRight(int key, int[] a, int lo, int hi){ 13 int mid = 0; 14 while(lo <= hi){ 15 mid = lo + (hi-lo)/2; 16 if(key > a[mid]) hi = mid -1; 17 else if(key < a[mid]) lo = mid +1; 18 else return mid; 19 } 20 return -1; 21 } 22 23 //worst case use 3lgn 24 public static int findNormal(int key, int[] a){ 25 int lo = 0; 26 int hi = a.length -1; 27 int mid = 0; 28 while(lo<=hi){ 29 mid = lo + (hi-lo)/2; 30 if(a[mid]==key) return mid; 31 else{ 32 if(mid==0 || mid==a.length-1) return -1; 33 if(a[mid]>a[mid-1] && a[mid]>a[mid+1]){//mid刚好取到最大数 34 break; 35 }else if(a[mid]>a[mid-1] && a[mid]<a[mid+1]){//mid取在增长序列 36 lo = mid +1; 37 }else{//mid取在衰减序列 38 hi = mid -1; 39 } 40 } 41 } 42 //循环结束，但没有return，说明key未找到，并且mid此时已经是取到最大数 43 int maxValIndex = mid; 44 if(key > a[maxValIndex]) return -1;//key大于a[i]中最大值，key肯定不在数组中 45 if(key < a[0] && key < a[a.length -1]) return -1; 46 int findValue = seachLeft(key,a,0,maxValIndex-1); 47 if(findValue == -1) 48 findValue = searchRight(key,a,maxValIndex+1,a.length-1); 49 50 return findValue; 51 } 52 53 public static int findFast(int key, int[] a){ 54 int lo = 0; 55 int hi = a.length -1; 56 int mid = 0; 57 int findValue = -1; 58 while(lo<=hi){ 59 mid = lo + (hi-lo)/2; 60 if(a[mid]==key) return mid; 61 else if(a[mid] > key){ 62 if(a[mid]>a[mid+1] && a[mid]>a[mid-1]){ 63 findValue = seachLeft(key,a,0,mid-1); 64 if(findValue != -1) return findValue; 65 else return searchRight(key,a,mid+1,a.length-1); 66 }else if(a[mid]>a[mid+1] && a[mid]<a[mid-1]){ 67 hi = mid-1; 68 }else{ 69 lo = mid +1; 70 } 71 }else{//说明key值可能在a[mid]的两边 72 findValue = seachLeft(key,a,0,mid-1); 73 if(findValue != -1) return findValue; 74 else return searchRight(key,a,mid+1,a.length-1); 75 } 76 } 77 return -1; 78 } 79 80 public static void main(String[] args){ 81 int[] a = {5,6,7,8,9,3,2,1,0}; 82 System.out.println(SearchBitonicArray.findNormal(4, a)); 83 System.out.println(SearchBitonicArray.findFast(4, a)); 84 } 85 }
Coursera Algorithms week1 Interview Questions: Search in a bitonic array
标签：break int nbsp hint without sea sequence ase rgs
原文地址：http://www.cnblogs.com/evasean/p/7208946.html
