要求
给一个没有重复元素的数组和目标值target，求距离该target最近的两个值，返回该距离(绝对值)
使用二分查找求目标值最近距离
定义两个指针left和high，并计算mid=left+(righ-left)/2,循环终止条件为left+1>=righ,循环终止说明目标值落在left和righ指向的两个值之间即nums[left]<=target<=nums[righ].即找到了距离目标值最近的两个值
private int binarySearch(int[] nums, int target)
{
Arrays.sort(nums);
int left = 0;
int righ = nums.length - 1;
while (left + 1 < righ)
{
int mid = left + (righ - left) / 2;
if (nums[mid] == target)
{
return 0;
}else if (nums[mid] < target)
{
left = mid;
}else
{
righ = mid;
}
}
return Math.min(Math.abs(target - nums[left]), Math.abs(target - nums[righ]));
}
