二分查找

待查找的序列是升序的，采用分治算法，时间复杂度为O(logN)

一个使用递归的版本

    public static int binarySearch1(int[] array, int toFind, int left, int right) {
        int mid = (right + left) / 2;
        // 基准情况
        if (right - left == 1) {
            if (array[left] == toFind)
                return left;
            return -1;
        }
        if (array[mid] > toFind) {
            return binarySearch1(array, toFind, left, mid);
        } else if (array[mid] < toFind) {
            return binarySearch1(array, toFind, mid, right);
        } else {
            return mid;
        }
    }

使用循环代替递归

    public static int binarySearch2(int[] array, int toFind) {
        int low = 0, high = array.length;
        while(low < high) {
            int mid = (low + high) / 2;
            if (array[mid] > toFind) {
                high = mid;
            } else if (array[mid] < toFind) {
                low = mid + 1;
            } else {
                return mid;
            }
        }
        return -1;
    }