Algorithm question check-in day58 - monotonic stack | 739. Daily temperature, 496. Next larger element I

739. Daily temperature - LeetCode

Status: Violence timeout, AC after checking the idea.

By continuously maintaining a monotonically increasing stack, the answer can be obtained in one traversal. The code is as follows:

class Solution {
public:
    vector<int> dailyTemperatures(vector<int>& temperatures) {
        int len = temperatures.size();
        vector<int> res(len, 0);
        stack<int> st;
        for(int i = 0; i < len; ++i){
            while(!st.empty() && temperatures[st.top()] < temperatures[i]){
                res[st.top()] = i - st.top();
                st.pop();
            }
            st.push(i);
        }
        return res;
    }
};

496. The next bigger element I - LeetCode

Status: AC after checking the train of thought.

The variation of the previous question mainly adds an unordered_map to match existing elements. The code is as follows:

class Solution {
public:
    vector<int> nextGreaterElement(vector<int>& nums1, vector<int>& nums2) {
        int len1 = nums1.size(), len2 = nums2.size();
        vector<int> res(len1, -1);
        unordered_map<int, int> unmap;
        for(int i = 0; i < len1; ++i){
            // key 是数组1的值，value是索引
            unmap[nums1[i]] = i;
        }

        stack<int> st;
        for(int j = 0; j < len2; ++j){
            while(!st.empty() && nums2[st.top()] < nums2[j]){
                if(unmap.count(nums2[st.top()]) > 0){
                    // nums1中存在
                    int index1 = unmap[nums2[st.top()]];
                    res[index1] = nums2[j];
                }
                st.pop();
            }
            st.push(j);
        }
        return res;
    }
};