文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

We have an integer array nums, where all the integers in nums are 0 or 1. You will not be given direct access to the array, instead, you will have an API ArrayReader which have the following functions:

• int query(int a, int b, int c, int d): where 0 <= a < b < c < d < ArrayReader.length().
  The function returns the distribution of the value of the 4 elements and returns:
  4 : if the values of the 4 elements are the same (0 or 1).
  2 : if three elements have a value equal to 0 and one element has value equal to 1 or vice versa.
  0 : if two element have a value equal to 0 and two elements have a value equal to 1.
• int length(): Returns the size of the array.
  You are allowed to call query() 2 * n times at most where n is equal to ArrayReader.length().

Return any index of the most frequent value in nums, in case of tie, return -1.

Follow up: What is the minimum number of calls needed to find the majority element?

Example 1: Input: nums = [0,0,1,0,1,1,1,1] Output: 5 Explanation: The following calls to the API reader.length() // returns 8 because there are 8 elements in the hidden array. reader.query(0,1,2,3) // returns 2 this is a query that compares the elements nums[0], nums[1], nums[2], nums[3] // Three elements have a value equal to 0 and one element has value equal to 1 or viceversa. reader.query(4,5,6,7) // returns 4 because nums[4], nums[5], nums[6], nums[7] have the same value. we can infer that the most frequent value is found in the last 4 elements. Index 2, 4, 6, 7 is also a correct answer.Example 2: Input: nums = [0,0,1,1,0] Output: 0Example 3: Input: nums = [1,0,1,0,1,0,1,0] Output: -1Constraints: 5 <= nums.length <= 10^5 0 <= nums[i] <= 1

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/guess-the-majority-in-a-hidden-array
著作權歸領扣網絡所有。商業轉載請聯系官方授權，非商業轉載請注明出處。

2. 解題

• 參考題解

/*** // This is the ArrayReader's API interface.* // You should not implement it, or speculate about its implementation* class ArrayReader {* public:* // Compares 4 different elements in the array* // return 4 if the values of the 4 elements are the same (0 or 1).* // return 2 if three elements have a value equal to 0 and one element has value equal to 1 or vice versa.* // return 0 : if two element have a value equal to 0 and two elements have a value equal to 1.* int query(int a, int b, int c, int d);** // Returns the length of the array* int length();* };*/class Solution { public:int guessMajority(ArrayReader &reader) {int n = reader.length();int start = reader.query(0,1,2,3);int g1 = 1, g2 = 0, idx1 = 0, idx2 = -1;//假設idx = 0 的數是第一類，其個數為 g1 = 1for(int i = 4; i < n; ++i){if(reader.query(1,2,3,i)==start)//0， i 是否是一類g1++;elseg2++, idx2 = i;}//0 與 4,5...n-1 是否是一類//還要確定1,2,3int q = reader.query(0,2,3,4);int p = reader.query(1,2,3,4);if(q == p)//0和1是一類g1++;elseg2++, idx2 = 1;q = reader.query(0,1,3,4);// p = reader.query(1,2,3,4);if(q == p)//0和2是否是一類g1++;elseg2++,idx2 = 2;q = reader.query(0,1,2,4);// p = reader.query(1,2,3,4);if(q == p)//0和3是否是一類g1++;elseg2++,idx2 = 3;if(g1 == g2) return -1;if(g1 > g2) return idx1;return idx2;} };

280 ms 31 MB

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