1、 由于是有序数组，log(n)的时间查找最小值的index

1.1 一定要注意当m、r、l三个位置相同值时，无法区分左右了，所以需要在l到r区间遍历找到最小元素。

2、 看最小值index的左右两侧，是否查找值落在 [l,i] 与 [i,r] 上，在对应区间上进行二分查找

 1 #include<iostream>
 2 using namespace std;
 3 int findminpos(int* a, int l, int r){
 4     int pos = l;
 5     int min = a[l];
 6     for(int i=l+1; i<=r; i++){
 7         if( min>a[i] ){
 8             min=a[i];
 9             pos=i;
10         }
11     }
12     return pos;
13 }
14 int findBound(int *a, int r){
15     if ( r==0 ) return 0;
16     int l = 0;
17     while(l<r){
18         if( r-l == 1 )break;
19         int m = (l+r)>>1;
20         cout << "search at: " << m << endl;
21         if( a[m] == a[r] && a[l] == a[m] ) return findminpos(a,l,r);
22         if( a[m] > a[r] ) l = m;
23         else r = m;
24     }
25     int p = l;
26     if ( a[l] > a[r] ) p = r;
27     return r;
28 }
29
30 bool find(int *a, int s, int arrlen){
31     if( !a ) return false;
32     int l = 0;
33     int r = arrlen-1;
34     int m = findBound(a, r);
35     cout << "start pos: " <<  m << endl;
36     if( s >= a[m] && s <= a[r] ) {
37         l = m;
38     } else {
39         r = m-1;
40     }
41     cout << "searching set [" << l << ","<< r << "]" << endl;
42     while ( l <= r ){
43         m = (l+r)>>1;
44         if( a[m] == s) return true;
45         if( a[m] > s ) r = m-1;
46         else l = m + 1;
47     }
48     return false;
49 }
50
51 int main(){
52     int a[] = {1,1,1,1,0,1,1};
53     if(find(a, 1, 7)) {
54         cout << "find" << endl;
55     } else {
56         cout << "not found" << endl;
57     }
58 }