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[LeetCode] 56. Merge Intervals #56

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grandyang opened this issue May 30, 2019 · 0 comments
Open

[LeetCode] 56. Merge Intervals #56

grandyang opened this issue May 30, 2019 · 0 comments

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@grandyang
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@grandyang grandyang commented May 30, 2019

 

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.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

 

这道和之前那道 Insert Interval 很类似,这次题目要求我们合并区间,之前那题明确了输入区间集是有序的,而这题没有,所以我们首先要做的就是给区间集排序,由于我们要排序的是个结构体,所以我们要定义自己的 comparator,才能用 sort 来排序,我们以 start 的值从小到大来排序,排完序我们就可以开始合并了,首先把第一个区间存入结果中,然后从第二个开始遍历区间集,如果结果中最后一个区间和遍历的当前区间无重叠,直接将当前区间存入结果中,如果有重叠,将结果中最后一个区间的 end 值更新为结果中最后一个区间的 end 和当前 end 值之中的较大值,然后继续遍历区间集,以此类推可以得到最终结果,代码如下:

 

解法一:

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        if (intervals.empty()) return {};
        sort(intervals.begin(), intervals.end());
        vector<vector<int>> res{intervals[0]};
        for (int i = 1; i < intervals.size(); ++i) {
            if (res.back()[1] < intervals[i][0]) {
                res.push_back(intervals[i]);
            } else {
                res.back()[1] = max(res.back()[1], intervals[i][1]);
            }
        }   
        return res;
    }
};

 

下面这种解法将起始位置和结束位置分别存到了两个不同的数组 starts 和 ends 中,然后分别进行排序,之后用两个指针i和j,初始化时分别指向 starts 和 ends 数组的首位置,然后如果i指向 starts 数组中的最后一个位置,或者当 starts 数组上 i+1 位置上的数字大于 ends 数组的i位置上的数时,此时说明区间已经不连续了,我们来看题目中的例子,排序后的 starts 和 ends 为:

starts:    1    2    8    15

ends:     3    6    10    18

红色为i的位置,蓝色为j的位置,那么此时 starts[i+1] 为8,ends[i] 为6,8大于6,所以此时不连续了,将区间 [starts[j], ends[i]],即 [1, 6] 加入结果 res 中,然后j赋值为 i+1 继续循环,参见代码如下:

 

解法二:

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        int n = intervals.size();
        vector<vector<int>> res;
        vector<int> starts, ends;
        for (int i = 0; i < n; ++i) {
            starts.push_back(intervals[i][0]);
            ends.push_back(intervals[i][1]);
        }
        sort(starts.begin(), starts.end());
        sort(ends.begin(), ends.end());
        for (int i = 0, j = 0; i < n; ++i) {
            if (i == n - 1 || starts[i + 1] > ends[i]) {
                res.push_back({starts[j], ends[i]});
                j = i + 1;
            }
        } 
        return res;
    }
};

 

这道题还有另一种解法,这个解法直接调用了之前那道题 Insert Interval 的函数,由于插入的过程中也有合并的操作,所以我们可以建立一个空的集合,然后把区间集的每一个区间当做一个新的区间插入结果中,也可以得到合并后的结果,那道题中的四种解法都可以在这里使用,但是没必要都列出来,这里只选了那道题中的解法二放到这里,代码如下:

 

解法三:

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int>> res;
        for (int i = 0; i < intervals.size(); ++i) {
            res = insert(res, intervals[i]);
        }
        return res;
    }
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int> newInterval) {
        vector<vector<int>> res;
        int n = intervals.size(), cur = 0;
        for (int i = 0; i < n; ++i) {
            if (intervals[i][1] < newInterval[0]) {
                res.push_back(intervals[i]);
                ++cur;
            } else if (intervals[i][0] > newInterval[1]) {
                res.push_back(intervals[i]);
            } else {
                newInterval[0] = min(newInterval[0], intervals[i][0]);
                newInterval[1] = max(newInterval[1], intervals[i][1]);
            }
        }
        res.insert(res.begin() + cur, newInterval);
        return res;
    }
};

 

Github 同步地址:

#56

 

类似题目:

Employee Free Time

Insert Interval

Meeting Rooms II

Meeting Rooms

Teemo Attacking

Add Bold Tag in String

Range Module

Partition Labels

Interval List Intersections

 

参考资料:

https://leetcode.com/problems/merge-intervals/

https://leetcode.com/problems/merge-intervals/discuss/21242/C++-10-line-solution.-easing-understanding

https://leetcode.com/problems/merge-intervals/discuss/21223/Beat-98-Java.-Sort-start-and-end-respectively

 

LeetCode All in One 题目讲解汇总(持续更新中...)

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