Example of Java using dichotomy to find and sort

To realize the dichotomy search, the array needs to be an orderly sequence. The dichotomy search is more efficient than the linear search: the more elements of the array, the more obvious the efficiency improvement. The efficiency of dichotomy search means that O (log2n) n is in the M-power range of 2, the maximum number of searches is m, log2n means that the M-power of 2 is equal to N, and the constant is omitted, Abbreviated as O (logn) if there is an ordered array of 200 elements, the maximum number of binary lookups is 2 ^ 7 = 128 and 2 ^ 8 = 256. It can be seen that the power of 7 is less than 200 and the power of 8 is included, so the maximum number of lookups is equal to 8