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剑指 Offer II 074. 合并区间

题目描述

以数组 intervals 表示若干个区间的集合，其中单个区间为 intervals[i] = [starti, endi] 。请你合并所有重叠的区间，并返回一个不重叠的区间数组，该数组需恰好覆盖输入中的所有区间。

 

示例 1：

输入：intervals = [[1,3],[2,6],[8,10],[15,18]]
输出：[[1,6],[8,10],[15,18]]
解释：区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].

示例 2：

输入：intervals = [[1,4],[4,5]]
输出：[[1,5]]
解释：区间 [1,4] 和 [4,5] 可被视为重叠区间。

 

提示：

  • 1 <= intervals.length <= 104
  • intervals[i].length == 2
  • 0 <= starti <= endi <= 104

 

注意：本题与主站 56 题相同： https://leetcode.cn/problems/merge-intervals/

解法

方法一：区间合并

区间合并，将所有存在交集的区间进行合并。方法是：先对区间按照左端点升序排列，然后遍历区间进行合并。

模板：

def merge(intervals):
    ans = []
    intervals.sort()
    st, ed = intervals[0]
    for s, e in intervals[1:]:
        if ed < s:
            ans.append([st, ed])
            st, ed = s, e
        else:
            ed = max(ed, e)
    ans.append([st, ed])
    return ans

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class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        intervals.sort()
        ans = []
        st, ed = intervals[0]
        for s, e in intervals[1:]:
            if ed < s:
                ans.append([st, ed])
                st, ed = s, e
            else:
                ed = max(ed, e)
        ans.append([st, ed])
        return ans

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class Solution {
    public int[][] merge(int[][] intervals) {
        Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
        int st = intervals[0][0], ed = intervals[0][1];
        List<int[]> ans = new ArrayList<>();
        for (int i = 1; i < intervals.length; ++i) {
            int s = intervals[i][0], e = intervals[i][1];
            if (ed < s) {
                ans.add(new int[] {st, ed});
                st = s;
                ed = e;
            } else {
                ed = Math.max(ed, e);
            }
        }
        ans.add(new int[] {st, ed});
        return ans.toArray(new int[ans.size()][]);
    }
}

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

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func merge(intervals [][]int) [][]int {
    sort.Slice(intervals, func(i, j int) bool {
        return intervals[i][0] < intervals[j][0]
    })
    st, ed := intervals[0][0], intervals[0][1]
    var ans [][]int
    for _, e := range intervals[1:] {
        if ed < e[0] {
            ans = append(ans, []int{st, ed})
            st, ed = e[0], e[1]
        } else if ed < e[1] {
            ed = e[1]
        }
    }
    ans = append(ans, []int{st, ed})
    return ans
}

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public class Solution {
    public int[][] Merge(int[][] intervals) {
        intervals = intervals.OrderBy(a => a[0]).ToArray();
        int st = intervals[0][0], ed = intervals[0][1];
        var ans = new List<int[]>();
        for (int i = 1; i < intervals.Length; ++i)
        {
            int s = intervals[i][0], e = intervals[i][1];
            if (ed < s)
            {
                ans.Add(new int[]{st, ed});
                st = s;
                ed = e;
            }
            else
            {
                ed = Math.Max(ed, e);
            }
        }
        ans.Add(new int[]{st, ed});
        return ans.ToArray();
    }
}

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class Solution {
    func merge(_ intervals: [[Int]]) -> [[Int]] {
        guard !intervals.isEmpty else { return [] }

        let intervals = intervals.sorted { $0[0] < $1[0] }
        var result: [[Int]] = []

        var currentInterval = intervals[0]
        for interval in intervals.dropFirst() {
            if currentInterval[1] < interval[0] {
                result.append(currentInterval)
                currentInterval = interval
            } else {
                currentInterval[1] = max(currentInterval[1], interval[1])
            }
        }
        result.append(currentInterval)

        return result
    }
}

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