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Twitter OA prepare: K-complementary pair

A non-empty zero-indexed array A consisting of N integers is given.A pair of integers (P, Q) is called K-complementary in array A if 0 ≤ P, Q < N and A[P] + A[Q] = K.                For example, consider array A such that:                                  A[0] =  1  A[1] = 8  A[2]= -3                  A[3] =  0  A[4] = 1  A[5]=  3                  A[6] = -2  A[7] = 4  A[8]=  5The following pairs are 6-complementary in array A: (0,8), (1,6), (4,8), (5,5), (6,1), (8,0), (8,4). For instance, the pair (4,8) is 6-complementary because A[4] + A[8] = 1 + 5 = 6.                Write a function:class Solution { public int solution(int K, int[] A); }that, given an integer K and a non-empty zero-indexed array A consisting of N integers, returns the number of K-complementary pairs in array A.                For example, given K = 6 and array A such that:                                   A[0] =  1  A[1] = 8  A[2]= -3.                   A[3] =  0  A[4] = 1  A[5]=  3.                   A[6] = -2  A[7] = 4  A[8]=  5the function should return 7, as explained above.Assume that:N is an integer within the range [1..50,000];K is an integer within the range [−2,147,483,648..2,147,483,647];each element of array A is an integer within the range [−2,147,483,648..2,147,483,647].Complexity:expected worst-case time complexity is O(N*log(N));expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).Elements of input arrays can be modified.

很像2sum的夹逼算法,需要sort一下。本身不难,但是tricky的地方在于允许同一个数组元素自己跟自己组成一个pair,比如上例中的[5, 5]。而且数组本身就允许值相等的元素存在,在计算pair时,算成不同的pair,比如数组是[3,3],K=6,这时的pair有[0, 0], [0, 1], [1, 0], [1, 1]4个。

这个case让这道本来不难的题一下子麻烦了许多。我的对应处理方法是:用HashMap记录每个元素出现次数,用一个变量res记录可行pair数。 像夹逼方法那样,一左一右两个pointer l、r 分别往中间走,如果左右元素和加起来等于K:

1. 如果 l == r, res = res + 1;

2. else, 如果A[l] == A[r], res = res + Math.pow(map.get(A[l]), 2); 加上A[l]出现次数的平方,比如[3, 3]这个例子,加上2^2 ==4, 再如[3, 3, 3], 加9

3. else, 即A[l] != A[r], res = res + 2 * map.get(A[l]) * map.get(A[r]); 比如[1, 1, 5], 加上4;[1, 1, 5, 5],加上8;[-2, 8], 加上2

然后就是之后左右pointer该怎么跳,我的处理方法是,跳出现次数那么多次,这样就不再重复处理这些出现过的数字

 1 public int KComplementary(int[] A, int K) { 2     if (A==null || A.length==0) return 0; 3     int res = 0; 4     int l = 0; 5     int r = A.length - 1; 6     HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); 7     for (int i=0; i<A.length; i++) { 8         if (map.containsKey(A[i])) { 9             map.put(A[i], map.get(A[i])+1);10         }11         else {12             map.put(A[i], 1);13         }14     }15     while (l <= r) {16         if (A[l] + A[r] == K) {17             if (l == r) res += 1;18             else if (A[l] == A[r]) {19                 res += Math.pow(map.get(A[l]), 2);20             }21             else {22                 res += 2 * map.get(A[l]) * map.get(A[r]);23             }24             l = l + map.get(A[l]);25             r = r - map.get(A[r]);26         }27         else if (A[l] + A[r] < K) {28             l = l + map.get(A[l]);29         }30         else {31             r = r - map.get(A[r]);32         }33     }  34     return res;35 }    

 

Twitter OA prepare: K-complementary pair