算法题之无序数组找到任一高点
假设某无序数组A，首尾元素都是0，其余元素值都大于零，且不重复。每一个数值都代表一个高度。
要求找出A[1] 到A[N-2]之间的任一高点（高点定义是：某点的值大于其前后元素值），要求时间复杂度为O(Log(N)).
比如数组{0,8,9,4,5,1,3,6,7,0},其中高点为9,5,7.
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思路：要求为时间复杂度为O(Log(N))，那么遍历数组肯定无法实现，借鉴二分查找的思路，判断每个点的趋势，某点上升趋势的话，则后面必然有高点，前面不一定有；下降趋势的话，相反。
1 #include <iostream> 2 using namespace std; 3 enum trend 4 { 5 up = 0, 6 down, 7 submit, 8 peak, 9 foot, 10 none 11 }; 12 //get the thrend by checking the previous and following numbers. 13 trend getThrend(int a[], int count){ 14 if(count == 3) 15 { 16 if(a[0] > a[1]) 17 { 18 if(a[1] > a[2]) 19 return down; 20 else 21 return foot; 22 } 23 else 24 { 25 if(a[1] > a[2]) 26 return peak; 27 else 28 return up; 29 } 30 } 31 else 32 return none; 33 } 34 //find the peak point by using binary searching 35 void findOneHighPoint(int a[], int begin, int end){ 36 int mid = (begin+end)/2; 37 if(getThrend(&a[mid-1], 3) == peak) 38 { 39 cout<<"find one peak "<<a[mid]<<endl; 40 return; 41 } 42 if(getThrend(&a[mid-1], 3) == up) 43 findOneHighPoint(a, mid+1, end); 44 else 45 findOneHighPoint(a, begin, mid-1); 46 } 47 48 //input a array, begin end with 0 49 void main(){ 50 int a[10] = {0,8,9,4,5,6,6,6,7,0}; 51 findOneHighPoint(a, 1, 8); 52 }
