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leetcode-题解
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  1. templates

Monotonic-stack

单调栈可以用来求在一个数组中,位置i上的数左右离它最近的且比i位置上的数大(或小)的第一个数。首先,如果要求最近且比位置i上的数大,那么就要维持一个从栈底到栈顶由大到小的状态。如果新进来的数大于栈顶的数,那么栈顶数x右边最大的数就是即将新进来的数。x左边的最大的数就是栈顶地下的数。(这适用于数组中没有重复的数字)。

如果有重复值,那么栈中就不存放下标了,存放一个数组(保存的是大小相同的元素下标),每次弹出时,左边最近比它的是它下边队列后的最后一个下标位置的元素。

``` c++ "数组中无重复元素" //单调栈结构,求解i位置上左右离他最近且比它小的值 struct MonotonicStack { stack mono; //返回一个二维数组,每一行表示该位置左右离它最近的数,0左1右 vector> GetNearLessNoRepeat(vector& nums) { if (nums.empty()) { return vector> {}; } //一定要先开辟空间,否则无法使用下标 vector> results(nums.size(),vector(2)); //处理每个位置上的元素 for (int i = 0; i < (int)nums.size(); ++i) { //单调栈,栈底到栈顶,由小到大 while (!mono.empty() && nums[mono.top()] > nums[i]) { //栈不空,且栈顶元素比即将入栈的元素大,破坏了有大到小的结构(从栈顶到栈底) int current = mono.top(); mono.pop(); results[current][1] = i; results[current][0] = mono.empty() == true ? -1 : mono.top(); } //栈不为空且栈顶元素小于nums[i] mono.push(i); } //如果栈不为空,还需结算处理 while (!mono.empty()) { int popIndex = mono.top(); mono.pop(); results[popIndex][1] = -1; results[popIndex][0] = mono.empty() == true ? -1 : mono.top(); } return std::move(results); }

``` c++ "数组有重复元素"
 vector<vector<int>> GetLessRepeat(vector<int>& nums) {
        if (nums.empty()) {
            return vector<vector<int>>{};
        }
        stack<vector<int>> mono;
        vector<vector<int>> results(nums.size(),vector<int>(2));

        for (int i = 0; i < (int)nums.size(); ++i) {
            //比是和第一个元素比,求最左时是最后一个元素
            while (!mono.empty() && nums[mono.top().front()] > nums[i]) {
                vector<int> current_sets = mono.top();
                mono.pop();
                //弹栈时需要求每个元素的最左最右值
                for (auto elem : current_sets) {
                    results[elem][1] = i;
                    results[elem][0] = mono.empty() == true ? -1 : mono.top().back();
                }
            }
            //插入时要看是和top相等还是比top大
            if (!mono.empty() && nums[mono.top().back()] == nums[i]) {
                mono.top().push_back(i);
            }
            //比栈顶大或者为空,那就直接压栈就完事了
            else {
                mono.push(vector<int>{i});
            }
        }
        //清零操作
        while (!mono.empty()) {
            auto temp = mono.top();
            mono.pop();
            for (auto elem : temp) {
                //如果栈不为空,可以先取栈顶元素,避免多次访问栈
                results[elem][1] = -1;
                results[elem][0] = mono.empty() == true ? -1 : mono.top().back();
            }
        }
        return std::move(results);
    }
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