C++ | Leetcode C++题解之第352题将数据流变为多个不想交区间

题目：

题解：

class SummaryRanges {
private:map<int, int> intervals;public:SummaryRanges() {}void addNum(int val) {// 找到 l1 最小的且满足 l1 > val 的区间 interval1 = [l1, r1]// 如果不存在这样的区间，interval1 为尾迭代器auto interval1 = intervals.upper_bound(val);// 找到 l0 最大的且满足 l0 <= val 的区间 interval0 = [l0, r0]// 在有序集合中，interval0 就是 interval1 的前一个区间// 如果不存在这样的区间，interval0 为尾迭代器auto interval0 = (interval1 == intervals.begin() ? intervals.end() : prev(interval1));if (interval0 != intervals.end() && interval0->first <= val && val <= interval0->second) {// 情况一return;}else {bool left_aside = (interval0 != intervals.end() && interval0->second + 1 == val);bool right_aside = (interval1 != intervals.end() && interval1->first - 1 == val);if (left_aside && right_aside) {// 情况四int left = interval0->first, right = interval1->second;intervals.erase(interval0);intervals.erase(interval1);intervals.emplace(left, right);}else if (left_aside) {// 情况二++interval0->second;}else if (right_aside) {// 情况三int right = interval1->second;intervals.erase(interval1);intervals.emplace(val, right);}else {// 情况五intervals.emplace(val, val);}}}vector<vector<int>> getIntervals() {vector<vector<int>> ans;for (const auto& [left, right]: intervals) {ans.push_back({left, right});}return ans;}
};