README.md

56. 合并区间

English Version

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

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

Python3

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

Java

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()][]);
    }
}

C++

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;
    }
};

Go

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
}

C#

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();
    }
}

TypeScript

function merge(intervals: number[][]): number[][] {
    intervals.sort((a, b) => a[0] - b[0]);
    let ans: number[][] = [];
    let index: number = -1;
    for (let interval of intervals) {
        if (index == -1 || ans[index][1] < interval[0]) {
            // 保留
            ans.push(interval);
            index++;
        } else {
            // 求交集
            ans[index][1] = Math.max(ans[index][1], interval[1]);
        }
    }
    return ans;
}

function merge(intervals: number[][]): number[][] {
    intervals.sort((a, b) => a[0] - b[0]);
    const n = intervals.length;
    const res = [];
    let i = 0;
    while (i < n) {
        let [l, r] = intervals[i];
        i++;
        while (i < n && r >= intervals[i][0]) {
            r = Math.max(r, intervals[i][1]);
            i++;
        }
        res.push([l, r]);
    }
    return res;
}

Rust

impl Solution {
    pub fn merge(mut intervals: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        intervals.sort_unstable_by(|a, b| a[0].cmp(&b[0]));
        let n = intervals.len();
        let mut res = vec![];
        let mut i = 0;
        while i < n {
            let l = intervals[i][0];
            let mut r = intervals[i][1];
            i += 1;
            while i < n && r >= intervals[i][0] {
                r = r.max(intervals[i][1]);
                i += 1;
            }
            res.push(vec![l, r]);
        }
        res
    }
}

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