这题是用分治算法中的折半查找，针对的是一个有序数组，把数组分而治之，把输入分成两个与原来相同的子问题，如果规模还太大，我们对这些子问题再做上述处理，直到这些子问题容易解决为止。这种算法是一种效率很高的算法，每一次循环，它都能将数组的规模减少到原来的半，其空间复杂度是O(log(n))次。-This question is divide and conquer algorithm binary search is for an ordered array, the array of divide and rule, the same as the original input is divided into two sub-problems, if the scale is too large, we do the above for these sub-problems processed until these sub-problems easier to solve so far. This algorithm is a very efficient algorithm, each time through the loop, it can reduce the size of the array to the original and a half, its space complexity is O (log (n)) times.