内容简介：Given a collection of intervals, merge all overlapping intervals.Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].Explanation: Intervals [1,4] and [4,5] are considered overlapping.
题目
Merge interval
Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: [[1,4],[4,5]] Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
Insert interval
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].
解法
Merge interval，首先需要对所有的区间进行排序，排序的准则：
- 先判断区间左侧是否关系：较小者 排序 在前
- 如果左边界相等，根据右边界决定关系，但是这里将右侧大的往前，其实不重要
然后就遍历数组，进行合并即可：
- 前面区间和当前区间没有交集，直接将前面区间放入结果
- 有交集，则进行合并
整体复杂度是： O(NlogN)
Insert interval: 这一题可以偷懒，利用 merge interval 来实现同样的目的。即：
- 将新区间插入，排序
- merge
但是这种做法没有充分利用原来区间就已经有序的条件。插入一个新区间，主要就是判断新区间在原来区间中的 位置。
[], [], [], —, [], [], [], — [], [] [], [], [—], [], [], [] — [], [] [], [], [], —, [], [], [—], [], [] [], [], [—], [], [], [—], [], []
上图中的 dash 线表示新区建的左侧和右侧，它与原区间列表存在上述四种关系。对于上述四种关系，我们对于 dash 线前和 dash 线后的区间可以直接复制到结果中，中间直接和新区间 merge 即可。
那么中关系怎么查找呢？可以利用原来数组就是有序的前提进行二分查找。整体复杂度在 O(N)
思路
- 排序
- 二分查找边界
代码
class Solution {
private:
typedef vector<int> V;
typedef vector<V> VV;
struct Comp {
bool operator()(vector<int>& lhs, vector<int>& rhs) {
if (lhs[0] != rhs[0]) return lhs[0] < rhs[0];
return lhs[1] < rhs[1];
}
};
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), Comp());
int size = intervals.size();
VV result;
if (size == 0) return result;
auto prev = intervals[0];
for (int i = 1; i < size; ++i) {
auto& a = intervals[i];
if (prev[1] < a[0]) {
result.push_back(prev);
prev = a;
} else {
// prev[1] >= a[0]
prev[1] = max(prev[1], a[1]);
}
}
result.push_back(prev);
return result;
}
};
class Solution {
private:
typedef vector<int> V;
typedef vector<V> VV;
struct Comp {
bool operator()(vector<int>& lhs, vector<int>& rhs) {
if (lhs[0] != rhs[0]) return lhs[0] < rhs[0];
return lhs[1] < rhs[1];
}
};
public:
vector<vector<int>> insert1(vector<vector<int>>& intervals, vector<int>& newInterval) {
intervals.push_back(newInterval);
sort(intervals.begin(), intervals.end(), Comp());
int size = intervals.size();
VV result;
auto prev = intervals[0];
for (int i = 1; i < size; ++i) {
auto& a = intervals[i];
if (prev[1] < a[0]) {
result.push_back(prev);
prev = a;
} else {
// prev[1] >= a[0]
prev[1] = max(prev[1], a[1]);
}
}
result.push_back(prev);
return result;
}
vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
VV result;
int size = intervals.size();
if (size == 0) {
result.push_back(newInterval);
return result;
}
// corner case
if (newInterval[1] < intervals[0][0]) {
result.push_back(newInterval);
copy(intervals.begin(), intervals.end(), back_inserter(result));
return result;
} else if (newInterval[0] > intervals[size-1][1]) {
copy(intervals.begin(), intervals.end(), back_inserter(result));
result.push_back(newInterval);
return result;
}
int low = 0;
int high = size;
int target = newInterval[0];
int pos_b = -1;
while (low < high) {
int mid = (high - low) / 2 + low;
auto& h = intervals[mid];
if (h[0] <= target && h[1] >= target) {
pos_b = mid; break;
} else if (h[1] < target) {
low = mid + 1;
} else if (h[0] > target) {
high = mid;
}
}
int low_b = low;
target = newInterval[1];
low = 0; high = size;
int pos_e = -1;
while (low < high) {
int mid = (high - low) / 2 + low;
auto& h = intervals[mid];
if (h[0] <= target && h[1] >= target) {
pos_e = mid; break;
} else if (h[1] < target) {
low = mid + 1;
} else if (h[0] > target) {
high = mid;
}
}
int low_e = low;
if (pos_e == -1 && pos_b == -1) {
auto beg = intervals.begin();
auto end = intervals.begin();
advance(end, low_b);
copy(beg, end, back_inserter(result));
result.push_back(newInterval);
end = intervals.end();
advance(beg, low_e);
copy(beg, end, back_inserter(result));
} else if (pos_b == -1) {
auto beg = intervals.begin();
auto end = intervals.begin();
advance(end, low_b);
copy(beg, end, back_inserter(result));
newInterval[1] = intervals[pos_e][1];
result.push_back(newInterval);
advance(beg, min(pos_e + 1, size));
end = intervals.end();
copy(beg, end, back_inserter(result));
} else if (pos_e == -1) {
auto beg = intervals.begin();
auto end = intervals.begin();
advance(end, pos_b);
copy(beg, end, back_inserter(result));
newInterval[0] = intervals[pos_b][0];
result.push_back(newInterval);
advance(beg, low_e);
end = intervals.end();
copy(beg, end, back_inserter(result));
} else {
auto beg = intervals.begin();
auto end = intervals.begin();
advance(end, pos_b);
copy(beg, end, back_inserter(result));
newInterval[0] = intervals[pos_b][0];
newInterval[1] = intervals[pos_e][1];
result.push_back(newInterval);
advance(beg, min(pos_e + 1, size));
end = intervals.end();
copy(beg, end, back_inserter(result));
}
return result;
}
};
本文完
.
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