365天挑战LeetCode1000题——Day 031 每日一题 + 二分查找 09

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真正写对「二分查找」，从来不在于我们把区间写成了「左闭右开」还是「左闭右闭」，而是 在于我们能够根据题意：得到某种单调性，和可以逐步缩小搜索规模的条件，进而准确地设计可以使得搜索区间缩小的条件。

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500提交纪念

剑指 Offer II 041. 滑动窗口的平均值

代码实现（首刷自解）

class MovingAverage {
public:
    /** Initialize your data structure here. */
    queue<int> q;
    int size;
    double sum = 0;

    MovingAverage(int size) {
        this->size = size;
    }
    
    double next(int val) {
        if (q.size() == size) {
            sum += val - q.front();
            q.push(val), q.pop();
        }
        else {
            sum += val;
            q.push(val);
        }
        return sum / q.size();
    }
};

/**
 * Your MovingAverage object will be instantiated and called as such:
 * MovingAverage* obj = new MovingAverage(size);
 * double param_1 = obj->next(val);
 */

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1838. 最高频元素的频数

代码实现（首刷自解）

class Solution {
public:
    int maxFrequency(vector<int>& nums, int k) {
        int n = nums.size();
        int l = 1, r = n;
        sort(nums.begin(), nums.end());
        while (l < r) {
            int m = (l + r + 1) >> 1;
            long long minNum = 0;
            long long num = 0;

            for (int i = 0; i < m - 1; i++) 
                num += nums[m - 1] - nums[i];
            minNum = num;
            for (int k = m; k < n; k++) {
                num += - (nums[k - 1] - nums[k - m]) +
                (long long) (m - 1) * (nums[k] - nums[k - 1]);
                minNum = min(minNum, num);
            }

            if (minNum <= k) l = m;
            else r = m - 1;
        }
        return l;
    }
};

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540. 有序数组中的单一元素

代码实现（首刷自解）

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int ans = 0;
        for (auto num : nums) ans ^= num;
        return ans;
    }
};