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LeetCode57 Insert Interval
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:
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]. （Hard）
分析：
首先可以采用merge interval的方法，先把区间填进去，然后排序，最后再调用merge，复杂度O(NlogN)
代码：
1 /** 2 * Definition for an interval. 3 * struct Interval { 4 * int start; 5 * int end; 6 * Interval() : start(0), end(0) {} 7 * Interval(int s, int e) : start(s), end(e) {} 8 * }; 9 */10 class Solution {11 private:12 static bool cmp (const Interval& I1, const Interval& I2) {13 return I1.start < I2.start;14 }15 vector<Interval> merge(vector<Interval>& intervals) {16 vector<Interval> result;17 if (intervals.size() == 0) {18 return result;19 }20 sort(intervals.begin(), intervals.end(), cmp);21 int left = intervals[0].start, right = intervals[0].end;22 for (int i = 1; i < intervals.size(); ++i) {23 if (intervals[i].start <= right) {24 right = max(right,intervals[i].end);25 }26 else {27 result.push_back(Interval(left,right));28 left = intervals[i].start;29 right = intervals[i].end;30 }31 }32 result.push_back(Interval(left,right));33 return result;34 }35 public:36 vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {37 vector<Interval> result;38 intervals.push_back(newInterval);39 sort(intervals.begin(),intervals.end(),cmp);40 result = merge(intervals);41 return result;42 }43 };
当然还有O(N)的算法可以优化。
分三步来做：
第一步，找到左侧不跟newInterval相交的区间添加到结果中；
第二步，找到所有和newInterval相交的区间并找到其左边界和右边界，然后建立新的interval添加到结果中；
第三部，找到右侧不跟newInterval相交的区间添加到结果中。
注意很多细节在里面可能会犯错
代码：
1 /** 2 * Definition for an interval. 3 * struct Interval { 4 * int start; 5 * int end; 6 * Interval() : start(0), end(0) {} 7 * Interval(int s, int e) : start(s), end(e) {} 8 * }; 9 */10 class Solution {11 public:12 vector<Interval> insert(vector<Interval>& intervals, Interval newInterval) {13 vector<Interval> result;14 if (intervals.size() == 0) {15 result.push_back(newInterval);16 return result;17 }18 int i = 0;19 while (i < intervals.size() && intervals[i].end < newInterval.start) {20 result.push_back(intervals[i++]);21 }22 int left = 0;23 if (i == intervals.size()) { 24 left = newInterval.start; 25 }26 else {27 left = min(newInterval.start,intervals[i].start);28 }29 while (i < intervals.size() && intervals[i].start <= newInterval.end) {30 i++;31 }32 int right = 0;33 if (i >= 1) {34 right = max(newInterval.end, intervals[i - 1].end);35 }36 else {37 right = newInterval.end;38 }39 result.push_back(Interval(left,right));40 while (i < intervals.size() ) {41 result.push_back(intervals[i++]);42 }43 return result;44 }45 };
LeetCode57 Insert Interval
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