题目:
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:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]]
Example 2:
Input: intervals =[[1,2],[3,5],[6,7],[8,10],[12,16]]
, newInterval =[4,8]
Output: [[1,2],[3,10],[12,16]] Explanation: Because the new interval[4,8]
overlaps with[3,5],[6,7],[8,10]
.
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
描述:
给出一个按照起点升序排列过的区间,区间之间无交叉,现要求插入一个区间,输出最终无交叉的区间序列
分析:
跟先前的那道题(点这里)基本类似,不过不需要考虑排序的问题,然后按照起点的升序,把需要插入的元素放到合适的位置,再使用上题的做法...
提交错了好几次,是被vector的insert函数给坑到了,还是不熟悉...
代码:(时间复杂度 O (n))
class Solution {
public:
vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
int index = findInsertIndex(intervals, newInterval);
intervals.insert(intervals.begin() + index, newInterval);
return merge(intervals);
}
int findInsertIndex(const vector<vector<int>>& intervals, const vector<int>& newInterval) {
int len = intervals.size();
for (int i = 0; i < len; ++ i) {
if (intervals[i][0] >= newInterval[0]) {
return i;
}
}
return len;
}
vector<vector<int>> merge(const vector<vector<int>>& intervals) {
if (intervals.size() < 2) {
return intervals;
}
vector<vector<int>> result;
result.push_back(intervals[0]);
for (int i = 1; i < intervals.size(); ++ i) {
int last_index = result.size() - 1;
if (result[last_index][1] >= intervals[i][0]) {
result[last_index][1] = max(result[last_index][1], intervals[i][1]);
} else {
result.push_back(intervals[i]);
}
}
return result;
}
};