有序 循环数组的二分查找

例如 10，11，12，13，14，1，2，3，4，5，6，7，8，9虽然整体上不是有序的，但是只是在某个节点进行了循环。在这样的数组中进行查找，可以使用二分查找法。但是需要判断选环节的位置。

int Find(int* source, int size, int num){if(NULL == source || size < 0){return -1;}if(1== size){if(num == source[0]){return 0;}else{return -1;}}if(2 == size){if(num == source[0]){return 0;}else if(num == source[1]){return 1;}else{return -1;}}if(source[0] <= num && source[size/2]> num){int ret = Find(source,size/2,num);if(-1 !=ret){return ret;}else{return -1;}}else if(source[size/2] <= num && source[size - 1]>= num){int ret = Find(source + size/2,(size + 1)/2,num);if(-1 !=ret){return ret + size/2;}else{return -1;}}else if(source[0] >= num && source[size/2]>= num && source[size/2 + 1] >= num && source[size - 1]>= num){if(source[0] <= source[size/2]){int ret = Find(source + size/2,(size + 1)/2 ,num);if(-1 !=ret){return ret + size/2;}else{return -1;}}else if(source[size/2] <= source[size - 1]){int ret = Find(source,size/2,num);if(-1 !=ret){return ret;}else{return -1;}}else{return -1;}}else if(source[0] <= num && source[size/2]<= num && source[size/2 + 1] <= num && source[size - 1]<= num){if(source[0] <= source[size/2]){int ret = Find(source + size/2,(size + 1)/2 ,num);if(-1 !=ret){return ret + size/2;}else{return -1;}}else if(source[size/2 + 1] <= source[size - 1]){int ret = Find(source,size/2,num);if(-1 !=ret){return ret;}else{return -1;}}else{return -1;}}else{return -1;}}