滑动窗口好题

【leetcode】1438. 绝对差不超过限制的最长连续子数组

给你一个整数数组 nums ，和一个表示限制的整数 limit，请你返回最长连续子数组的长度，该子数组中的任意两个元素之间的绝对差必须小于或者等于 limit 。

如果不存在满足条件的子数组，则返回 0 。

很好的一道关于滑动窗口的题目，与上学期写TCP所用的思想有类似之处，重点在于用一个set进行查询最大和最小值（rbegin,begin）以及删除相关元素(erase(set.find(num)))的优化

 class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        multiset<int> window;
        int left = 0, right = 0, ans = 0;
        while(right < nums.size()) {
            window.insert(nums[right]);
            while(*window.rbegin() - *window.begin() > limit) { //if > : move the window
                window.erase(window.find(nums[left]));
                left++;
            }
            ans = max(ans, right - left + 1);
            right++;
        }
        return ans;
    }
};

两个解法（后者比前者时间复杂度更低，相当于另外使用一个循环求最大增量，类似于递推，后一状态等于减去首部加上尾部）

//方法一
class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int X) {
        int ans = 0;
        for(int i = 0; i < grumpy.size(); ++i) {
            if(grumpy[i] == 0) {
                ans += customers[i];
            }
        }
        int left = 0, right = X;
        int ans0 = ans;
        while(right <= customers.size()) {
            int temp = ans0;
            for(int i = left; i < right; ++i) {
                if(grumpy[i] == 1) {
                    temp += customers[i];
                }
            }
            ans = max(temp, ans);
            left++;
            right++;
        }
        return ans;
    }
};

//方法二
class Solution {
public:
    int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int X) {
        int total = 0;
        int n = customers.size();
        for (int i = 0; i < n; i++) {
            if (grumpy[i] == 0) {
                total += customers[i];
            }
        }
        int increase = 0;
        for (int i = 0; i < X; i++) {
            increase += customers[i] * grumpy[i];
        }
        int maxIncrease = increase;
        for (int i = X; i < n; i++) {
            increase = increase - customers[i - X] * grumpy[i - X] + customers[i] * grumpy[i];
            maxIncrease = max(maxIncrease, increase);
        }
        return total + maxIncrease;
    }
};