The Smallest Difference

Given two array of integers(the first array is array A, the second array is array B), 
now we are going to find a element in array A which is A[i], and another element in array B which is B[j],
so that the difference between A[i] and B[j] (|A[i] - B[j]|) is as small as possible,
return their smallest difference.

两个数组两个指针

两个数组的题常sort, 再通过两指针遍历

public int smallestDifference(int[] A, int[] B) {
        // write your code here
        if (A == null || B == null || A.length == 0 || B.length == 0) {
            return -1;
        }
        Arrays.sort(A);
        Arrays.sort(B);
        int i = 0, j = 0;
        int ans = Integer.MAX_VALUE;
        while (i < A.length && j < B.length) {
            if (ans == 0) {
                return ans;
            } else {
                ans = Math.min(ans, Math.abs(A[i] - B[j]));
            }
            if (A[i] < B[j]) {
                i++;
            } else {
                j++;
            }
            
        }
        return ans;
        
    }

 常常与0 的差的正负来判断,  来判断指针的走位, 此处直接判断大小更方便

 if (A[i] < B[j]) {
                i++;

 

原文地址:https://www.cnblogs.com/apanda009/p/7261557.html