algorithm 1.4.18

题目：

Local minimum of an array.Write a program that, given an arraya[]of N distinct integers, finds alocal minimum: an indexisuch that a[i-1] < a[i] < a[i+1].Your program should use ~2lgNcompares in the worst case

书中给出的解决思路是:

Answer: Examine the middle valuea[N/2]and its two neighbors a[N/2 - 1] and a[N/2 + 1]. Ifa[N/2]is a local minimum, stop; otherwise search in the half with the smaller neighbor

这个题目考虑了很久，也没有想明白，在stackoverflow里面看了一些回答之后，觉得这个题目给的条件是有问题的，或者说不充分的；

有几个点需要限制：

1.所有的元素不能有相同的，如果有相同的话，那么递归条件就需要进行重新处理；

思路：

array[from-to]，

1、两个端点array[from] < array[from+1] ，或者aray[to] < array[to-1]，是极小值；

2、如果array只有一个值，这种情况可以认为有极小值，也可以认为没有；

3、进入递归条件时，array满足下面的条件 array[from] > array[from+1],array[to] > array[to-1];

mid = (from+to)/2;

IF array[mid]  > array[mid+1]  THEN find in [mid, to]; 

IF array[mid] > array[mid-1] THEN find in [from ,mid];

IF array[mid] < array[mid+1] && array[mid-1] > array[mid] RETURN  mid;

ELSE break;

根据上面的算法，java实现如下：

package com.frozenxia.algorithm.basic.exercises;public class LocalMinimun1418 {public static void main(String[] args) {int arry[] = { 9, 8, 7, 3, 10 };System.out.println(new LocalMinimun1418().findLocaMinimal(arry));}int findLocaMinimal(int arry[]) {if (arry == null)return -1;int lenth = arry.length;if (lenth == 0)return -1;if (lenth == 1)return 0;if (lenth == 2)return arry[0] < arry[1] ? 0 : -1;if (arry[0] < arry[1])return 0;if (arry[lenth - 1] < arry[lenth - 2])return lenth - 1;int left = 0;int right = lenth - 1;while (left < right) {if (right - left < 2)break;int mid = (left + right) / 2;System.out.println(mid);if (arry[mid] > arry[mid - 1]) {right = mid;} else if (arry[mid] > arry[mid + 1]) {left = mid;} else if (arry[mid] < arry[mid - 1] && arry[mid] < arry[mid + 1]) {return mid;} else {break;}}return -1;}}