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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]= 5 The 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]= 5 the 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
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原文地址：http://www.cnblogs.com/EdwardLiu/p/4200218.html
