#面试编程题#一个有序数组（从小到大排列），数组中的数据有正有负，求这个数组中的最小绝对值。

BinarySearchAbsMin

public class BinarySearchAbsMin {
  public static void main(String[] args) {
		int[] ints = { -3213, -232, -453, -5, -3, -2, 3, 8, 834, 2342, 999 };

		int min = searchAbsMin(ints, 0, ints.length);
		System.out.println(min);
	}

	// 二分查找
	private static int searchAbsMin(int[] ints, int start, int end) {
		// TODO Auto-generated method stub

		int size = start+end ;
		
		if(size==2){
			return Math.min(Math.abs(ints[start]), Math.abs(ints[start +1])); 
		}
		if(size==1){
			return ints[start] ;
		}
		
		int mid = size / 2;
		
		// 如果正负数,那么就说明是这两之中,0 也比一下吧
		if (ints[mid] >= 0 && ints[mid - 1] <= 0) {
			return Math.min(ints[mid], -ints[mid - 1]);
		}

		// 如果都大于0.找左面
		if (ints[mid] > 0 && ints[mid - 1] > 0) {
			return searchAbsMin(ints, start, mid);
		}

		// 如果都小于0.找右面
		if (ints[mid] < 0 && ints[mid - 1] < 0) {
			return searchAbsMin(ints, mid, end);
		}

		return 0;
	}
}

/我写了一个非递归的：

int searchAbsMin(int arr[],int len){

int lt=0,rt=len-1;
while(lt<rt){
    if(arr[lt]>=0){
        return lt;
    }
    if(arr[rt]<=0){
        return rt;
    }

    if(lt==rt-1){
        return getAbs(arr[lt])<getAbs(arr[rt])?lt:rt;
    }

    int mid=(lt+rt)>>1;
    if(arr[mid]<=0){
        if(arr[mid+1]>=0){
            rt=mid;
        }
        else{
            lt=mid+1;
        }
    }
    else{
        rt=mid;
    }
}

}

int close_key_search(int* a, int begin, int end, int key)
{
if(a[begin] >= key)return begin;
if(a[end] <= key)return end;
return close_search(a, begin, end, key);
}
int close_search(int* a, int begin, int end, int key)
{
int mid = (begin+end)/2;

if(1 == (end - begin)) 
{
    if((key - a[begin]) > (a[end] - key))return end;
    else return begin;
}

if(a[mid] == key)return mid;
else if(a[mid] > key) return close_search(a, begin, mid-1, key);
else return close_search(a, mid+1, end, key);

}