LintCode-最小差

给定数组 A = [3,4,6,7]， B = [2,3,8,9]，返回 0。

时间复杂度 O(n log n)

分析：看到时间复杂度就想到遍历A数组然后对B数组进行二分查找，二分查找可以手写，也可以直接调用STL。

代码：

class Solution {
public:
    /**
     * @param A, B: Two integer arrays.
     * @return: Their smallest difference.
     */
    int smallestDifference(vector<int> &A, vector<int> &B) {
        // write your code here
        int ret = INT_MAX;
        sort(B.begin(),B.end());
        for(int i=0;i<A.size();i++)
        {
            int diff = lower_bound(B.begin(),B.end(),A[i])-B.begin();
            ret = min(ret,abs(A[i]-B[diff]));
            if(diff>0)
                ret = min(ret,abs(A[i]-B[diff-1]));
            if(diff+1<B.size())
                ret = min(ret,abs(A[i]-B[diff+1]));
        }
        return ret;
    }
};