57. Insert Interval , STL的使用
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Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Given intervals [1,3],[6,9]
, insert and merge [2,5]
in as [1,5],[6,9]
.
Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16]
, insert and merge [4,9]
in as [1,2],[3,10],[12,16]
.
This is because the new interval [4,9]
overlaps with [3,5],[6,7],[8,10]
.
/** * Definition for an interval. * struct Interval { * int start; * int end; * Interval() : start(0), end(0) {} * Interval(int s, int e) : start(s), end(e) {} * }; */class Solution {public: vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) { auto compare = [] (const Interval &intv1, const Interval &intv2) { return intv1.end < intv2.start; };//!!!!!!!!!注意了 不是return intv1.start < intv2.start; /*例如,A{1, 3} B{2,5}根据自定义的比较函数有:A.end > B.start 为false B.end > A.start 为false 故:A = B; 其意义是:如果A B重叠,那么A = B所以,根据lower_bounder 二分查找后,如果找到的最左位置为index,返回index*/ auto range = equal_range(intervals.begin(), intervals.end(), newInterval, compare); auto itr1 = range.first, itr2 = range.second; if (itr1 == itr2) { intervals.insert(itr1, newInterval); } else { itr2--; itr2->start = min(newInterval.start, itr1->start); itr2->end = max(newInterval.end, itr2->end); intervals.erase(itr1, itr2);//注意了!!! 这里erase的是[itr1, itr2) 切记 } return intervals; }};
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