Lesson 8 question 2 EquiLeader

源地址

https://app.codility.com/programmers/lessons/8-leader/equi_leader/

EquiLeader

給定一個含有N個整數的非空數組A.
一個 equi leader 指的是下標S ,數組A被S ( 0 ≤ S < N − 1) 分割爲兩個數組,這兩個數組擁有相同的Leader.
比如：

	A[0] = 4
    A[1] = 3
    A[2] = 4
    A[3] = 4
    A[4] = 4
    A[5] = 2

有兩個 equi leader

• 0, 因爲 (4) 和 (3, 4, 4, 4, 2) 有相同的 leader 4.
• 2, 因爲 (4, 3, 4) 和 (4, 4, 2) 有相同的 leader 4.

寫一個函數,給定一個數組包含N個整數的非空數組A,返回 equi leader的個數.

	class Solution { 
		public int solution(int[] A); 
	}

比如

	A[0] = 4
    A[1] = 3
    A[2] = 4
    A[3] = 4
    A[4] = 4
    A[5] = 2

函數返回2.
假定：

• N是範圍在 [1…100,000]的整數
• A中的每個元素都是範圍在 [−1,000,000,000…1,000,000,000]的整數

第一步

如果把一個數組分爲兩個子數組,兩個字數組的Leader相等,那麼這子Leader一定和原數組的Leader相等.
那麼可以先求出數組A的Leader,然後依次判斷各個子數組是否含有相同值的Leader.

	public int solution(int[] A) {
        int count = 0;
        Integer leader = getLeader(A);
        if (leader == null) {
            return count;
        }

        int N = A.length;
        for (int i = 0; i < N - 1; i++) {

            int countLeft = 0;
            for (int j = 0; j <= i; j++) {
                if (A[j] == leader.intValue()) {
                    countLeft++;
                }
            }
            if (countLeft <= (i + 1) / 2)
                continue;

            int countRight = 0;
            for (int j = i + 1; j < N; j++) {
                if (A[j] == leader.intValue()) {
                    countRight++;
                }
            }
            if (countRight > (N - i - 1) / 2)
                count++;
        }

        return count;
    }

    Integer getLeader(int[] A) {
        int N = A.length;
        LinkedList<Integer> stack = new LinkedList<Integer>();
        for (int i = 0; i < N; i++) {
            if (stack.size() == 0) {
                stack.add(A[i]);
            } else {
                if (stack.getLast() != A[i]) {
                    stack.removeLast();
                } else {
                    stack.addLast(A[i]);
                }
            }
        }
        if (stack.size() == 0)
            return null;
        int candidate = stack.getLast();
        int count = 0;
        for (int i = 0; i < N; i++) {
            if (A[i] == candidate) {
                count++;
            }
        }
        return count > N / 2 ? candidate : null;
    }

優化一下

上面的solution中有重複計算的部分.在計算子數組中candidate的個數的時候,不用每一次都從0到N-1計算一邊,只有第一次需要遍歷全部,後面每次移動index的時候判斷當前元素是否等於candidate,然後加上或者建議1 即可.

	public int solution(int[] A) {
        int count = 0;
        Integer leader = getLeader(A);
        if (leader == null) {
            return count;
        }

        int N = A.length;
        int countLeft = 0;
        int countRight = 0;

        // calculate the count of leader where index = 0
        if (A[0] == leader.intValue())
            countLeft++;
        for (int i = 1; i < N; i++) {
            if (A[i] == leader.intValue())
                countRight++;
        }
        if (countLeft > 0 && countRight > (N - 1) / 2) {
            count++;
        }

        // move the index from 0 to N-1 and calculate the count of leader
        for (int i = 1; i < N - 1; i++) {
            if (A[i] == leader.intValue()) {
                countLeft++;
                countRight--;
            }
            if (countLeft > (i + 1) / 2 && countRight > (N - i - 1) / 2) {
                count++;
            }
        }

        return count;
    }

    Integer getLeader(int[] A) {
        int N = A.length;
        LinkedList<Integer> stack = new LinkedList<Integer>();
        for (int i = 0; i < N; i++) {
            if (stack.size() == 0) {
                stack.add(A[i]);
            } else {
                if (stack.getLast() != A[i]) {
                    stack.removeLast();
                } else {
                    stack.addLast(A[i]);
                }
            }
        }
        if (stack.size() == 0)
            return null;
        int candidate = stack.getLast();
        int count = 0;
        for (int i = 0; i < N; i++) {
            if (A[i] == candidate) {
                count++;
            }
        }
        return count > N / 2 ? candidate : null;
    }