Closest Number in Sorted Array

给出一个target，找出数组中与这个值最接近的值

  public int closestNumber(int[] A, int target) {
        // Write your code here
        if(A == null || A.length == 0){
            return -1;
        }
        int start = 0;
        int end = A.length - 1;

        while(start + 1 < end){
            int mid = start + (end - start)/2;
            if( A[mid] < target){
                start = mid;
            }else if (A[mid] == target){
                return mid;
            }else{
                end = mid;
            }
        }

        if(A[start] <= target && A[end] >= target){
            return Math.abs(A[start] - target) > Math.abs(A[end] - target)?end:start;
        }else if(A[end] < target){
            return end;
        }else{
            return start;           
        }
    }

K Closest Numbers In Sorted Array

find index, left pointer, right pointer and 

```

    public int[] kClosestNumbers(int[] A, int target, int k) {
    int[] result = new int[k];
    if ( A == null || A.length == 0 || k == 0 ) return result;
    int left = findIndex(A, target) - 1;
    int right =findIndex(A, target) + 1;
    result[0] = A[findIndex(A, target)];
    int index = 1;
    while (left >= 0 && right <= A.length - 1&& index < k ){
        if ( Math.abs( A[left] - target) <= Math.abs(A[right] - target)){
            result[index] = A[left];
            left--;
        }else{
            result[index] = A[right];
            right++;               
        }
        index++;
    }
    if ( index == k) return result;
    if ( left < 0 && right < A.length){
        while(right < A.length && index < k){
            result[index] = A[right];
            right++;
            index++;
        }
        return result;
    }else if ( right >= A.length && left >= 0 ){
        while(left >= 0 && index < k){
            result[index] = A[left];
            left--;
            index++;
        }
        return result;
    }
    return result;
}

public int findIndex( int[] A, int target){
    int start = 0;
    int end = A.length - 1;
    while ( start + 1 < end){
        int mid = start + (end - start) / 2;
        if ( A[mid] == target ) return mid;
        else if ( A[mid] < target){
            start = mid;
        }
        else {
            end = mid;
        }
    }
    if ( A[start] >= target) return start;
    else if ( A[end] <= target) return end;
    else return Math.abs( A[start] - target) > Math.abs(A[end] - target)? end:start;
}

```