Binary Search汇总

Binary Search：二分查找

实现方式：recursive or iterative

注意问题，终止循环条件，移动start，end时边界值，start = mid，end = mid

Template：

 

1. 在排序的数组中找A[i] == i的index，有重复元素存在. (cc150 Question 9.3)

因为有重复元素，所以例如：

A[mid] < mid, 不仅要search right,左边也可能出现A[i] == i

所以两边都需要Search，但是可以排除一些点

Left： start --> min{ A[mid], mid -1 } 

Right: max{ mid + 1, A[mid]} ----> end

 

/*
     * For question 2 elements are not distinct
     *search both left and right but can shrink searh area
     */
    public static int magicFast(int[] arr, int start, int end){
        if(end < start || start <  0 || end >= arr.length){
            return -1;
        }
        int midIndex = (start + end)/2;
        int midValue = arr[midIndex];
        
        if(midIndex == midValue) return midIndex;
        
        //search left
        int leftIndex = Math.min(midValue, midIndex - 1);
        int left = magicFast(arr, start, leftIndex);
        if(left >= 0) return left;
        
        //search right
        int rightIndex = Math.max(midValue, midIndex + 1);
        int right = magicFast(arr, rightIndex, end);
        //no need to check
        
        return right;
    }