comments	difficulty	edit_url	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0057.Insert%20Interval/README.md	数组

57. 插入区间

English Version

题目描述

给你一个 无重叠的 ，按照区间起始端点排序的区间列表 intervals，其中 intervals[i] = [starti, endi] 表示第 i 个区间的开始和结束，并且 intervals 按照 starti 升序排列。同样给定一个区间 newInterval = [start, end] 表示另一个区间的开始和结束。

在 intervals 中插入区间 newInterval，使得 intervals 依然按照 starti 升序排列，且区间之间不重叠（如果有必要的话，可以合并区间）。

返回插入之后的 intervals。

注意 你不需要原地修改 intervals。你可以创建一个新数组然后返回它。

 

示例 1：

输入：intervals = [[1,3],[6,9]], newInterval = [2,5]
输出：[[1,5],[6,9]]

示例 2：

输入：intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
输出：[[1,2],[3,10],[12,16]]
解释：这是因为新的区间 [4,8] 与 [3,5],[6,7],[8,10] 重叠。

 

提示：

• 0 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 105
• intervals 根据 starti 按 升序 排列
• newInterval.length == 2
• 0 <= start <= end <= 105

解法

方法一：排序 + 区间合并

我们可以先将新区间 newInterval 加入到区间列表 intervals 中，然后按照区间合并的常规方法进行合并。

时间复杂度 $O(n\times \log n)$，空间复杂度 $O(n)$。其中 $n$ 是区间的数量。

Python3

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

        intervals.append(newInterval)
        return merge(intervals)

Java

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        int[][] newIntervals = new int[intervals.length + 1][2];
        for (int i = 0; i < intervals.length; ++i) {
            newIntervals[i] = intervals[i];
        }
        newIntervals[intervals.length] = newInterval;
        return merge(newIntervals);
    }

    private int[][] merge(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        List<int[]> ans = new ArrayList<>();
        ans.add(intervals[0]);
        for (int i = 1; i < intervals.length; ++i) {
            int s = intervals[i][0], e = intervals[i][1];
            if (ans.get(ans.size() - 1)[1] < s) {
                ans.add(intervals[i]);
            } else {
                ans.get(ans.size() - 1)[1] = Math.max(ans.get(ans.size() - 1)[1], e);
            }
        }
        return ans.toArray(new int[ans.size()][]);
    }
}

C++

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        intervals.emplace_back(newInterval);
        return merge(intervals);
    }

    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end());
        vector<vector<int>> ans;
        ans.emplace_back(intervals[0]);
        for (int i = 1; i < intervals.size(); ++i) {
            if (ans.back()[1] < intervals[i][0]) {
                ans.emplace_back(intervals[i]);
            } else {
                ans.back()[1] = max(ans.back()[1], intervals[i][1]);
            }
        }
        return ans;
    }
};

Go

func insert(intervals [][]int, newInterval []int) [][]int {
	merge := func(intervals [][]int) (ans [][]int) {
		sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })
		ans = append(ans, intervals[0])
		for _, e := range intervals[1:] {
			if ans[len(ans)-1][1] < e[0] {
				ans = append(ans, e)
			} else {
				ans[len(ans)-1][1] = max(ans[len(ans)-1][1], e[1])
			}
		}
		return
	}
	intervals = append(intervals, newInterval)
	return merge(intervals)
}

TypeScript

function insert(intervals: number[][], newInterval: number[]): number[][] {
    const merge = (intervals: number[][]): number[][] => {
        intervals.sort((a, b) => a[0] - b[0]);
        const ans: number[][] = [intervals[0]];
        for (let i = 1; i < intervals.length; ++i) {
            if (ans.at(-1)[1] < intervals[i][0]) {
                ans.push(intervals[i]);
            } else {
                ans.at(-1)[1] = Math.max(ans.at(-1)[1], intervals[i][1]);
            }
        }
        return ans;
    };

    intervals.push(newInterval);
    return merge(intervals);
}

Rust

impl Solution {
    pub fn insert(intervals: Vec<Vec<i32>>, new_interval: Vec<i32>) -> Vec<Vec<i32>> {
        let mut merged_intervals = intervals.clone();
        merged_intervals.push(vec![new_interval[0], new_interval[1]]);
        // sort by elem[0]
        merged_intervals.sort_by_key(|elem| elem[0]);
        // merge interval
        let mut result = vec![];

        for interval in merged_intervals {
            if result.is_empty() {
                result.push(interval);
                continue;
            }

            let last_elem = result.last_mut().unwrap();
            if interval[0] > last_elem[1] {
                result.push(interval);
            } else {
                last_elem[1] = last_elem[1].max(interval[1]);
            }
        }
        result
    }
}

C#

public class Solution {
    public int[][] Insert(int[][] intervals, int[] newInterval) {
        int[][] newIntervals = new int[intervals.Length + 1][];
        for (int i = 0; i < intervals.Length; ++i) {
            newIntervals[i] = intervals[i];
        }
        newIntervals[intervals.Length] = newInterval;
        return Merge(newIntervals);
    }

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

方法二：一次遍历

我们可以遍历区间列表 intervals，记当前区间为 interval，对于每个区间有三种情况：

• 当前区间在新区间的右侧，即 $newInterval[1]<interval[0]$，此时如果新区间还没有被加入，那么将新区间加入到答案中，然后将当前区间加入到答案中。
• 当前区间在新区间的左侧，即 $interval[1]<newInterval[0]$，此时将当前区间加入到答案中。
• 否则，说明当前区间与新区间有交集，我们取当前区间的左端点和新区间的左端点的最小值，以及当前区间的右端点和新区间的右端点的最大值，作为新区间的左右端点，然后继续遍历区间列表。

遍历结束，如果新区间还没有被加入，那么将新区间加入到答案中。

时间复杂度 $O(n)$，其中 $n$ 是区间的数量。忽略答案数组的空间消耗，空间复杂度 $O(1)$。

Python3

class Solution:
    def insert(
        self, intervals: List[List[int]], newInterval: List[int]
    ) -> List[List[int]]:
        st, ed = newInterval
        ans = []
        insert = False
        for s, e in intervals:
            if ed < s:
                if not insert:
                    ans.append([st, ed])
                    insert = True
                ans.append([s, e])
            elif e < st:
                ans.append([s, e])
            else:
                st = min(st, s)
                ed = max(ed, e)
        if not insert:
            ans.append([st, ed])
        return ans

Java

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> ans = new ArrayList<>();
        int st = newInterval[0], ed = newInterval[1];
        boolean insert = false;
        for (int[] interval : intervals) {
            int s = interval[0], e = interval[1];
            if (ed < s) {
                if (!insert) {
                    ans.add(new int[] {st, ed});
                    insert = true;
                }
                ans.add(interval);
            } else if (e < st) {
                ans.add(interval);
            } else {
                st = Math.min(st, s);
                ed = Math.max(ed, e);
            }
        }
        if (!insert) {
            ans.add(new int[] {st, ed});
        }
        return ans.toArray(new int[ans.size()][]);
    }
}

C++

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        vector<vector<int>> ans;
        int st = newInterval[0], ed = newInterval[1];
        bool insert = false;
        for (auto& interval : intervals) {
            int s = interval[0], e = interval[1];
            if (ed < s) {
                if (!insert) {
                    ans.push_back({st, ed});
                    insert = true;
                }
                ans.push_back(interval);
            } else if (e < st) {
                ans.push_back(interval);
            } else {
                st = min(st, s);
                ed = max(ed, e);
            }
        }
        if (!insert) {
            ans.push_back({st, ed});
        }
        return ans;
    }
};

Go

func insert(intervals [][]int, newInterval []int) (ans [][]int) {
	st, ed := newInterval[0], newInterval[1]
	insert := false
	for _, interval := range intervals {
		s, e := interval[0], interval[1]
		if ed < s {
			if !insert {
				ans = append(ans, []int{st, ed})
				insert = true
			}
			ans = append(ans, interval)
		} else if e < st {
			ans = append(ans, interval)
		} else {
			st = min(st, s)
			ed = max(ed, e)
		}
	}
	if !insert {
		ans = append(ans, []int{st, ed})
	}
	return
}

TypeScript

function insert(intervals: number[][], newInterval: number[]): number[][] {
    let [st, ed] = newInterval;
    const ans: number[][] = [];
    let insert = false;
    for (const [s, e] of intervals) {
        if (ed < s) {
            if (!insert) {
                ans.push([st, ed]);
                insert = true;
            }
            ans.push([s, e]);
        } else if (e < st) {
            ans.push([s, e]);
        } else {
            st = Math.min(st, s);
            ed = Math.max(ed, e);
        }
    }
    if (!insert) {
        ans.push([st, ed]);
    }
    return ans;
}

Rust

impl Solution {
    pub fn insert(intervals: Vec<Vec<i32>>, new_interval: Vec<i32>) -> Vec<Vec<i32>> {
        let mut inserted = false;
        let mut result = vec![];

        let (mut start, mut end) = (new_interval[0], new_interval[1]);
        for iter in intervals.iter() {
            let (cur_st, cur_ed) = (iter[0], iter[1]);
            if cur_ed < start {
                result.push(vec![cur_st, cur_ed]);
            } else if cur_st > end {
                if !inserted {
                    inserted = true;
                    result.push(vec![start, end]);
                }
                result.push(vec![cur_st, cur_ed]);
            } else {
                start = std::cmp::min(start, cur_st);
                end = std::cmp::max(end, cur_ed);
            }
        }

        if !inserted {
            result.push(vec![start, end]);
        }
        result
    }
}

C#

public class Solution {
    public int[][] Insert(int[][] intervals, int[] newInterval) {
        var ans = new List<int[]>();
        int st = newInterval[0], ed = newInterval[1];
        bool insert = false;
        foreach (var interval in intervals) {
            int s = interval[0], e = interval[1];
            if (ed < s) {
                if (!insert) {
                    ans.Add(new int[]{st, ed});
                    insert = true;
                }
                ans.Add(interval);
            } else if (st > e) {
                ans.Add(interval);
            } else {
                st = Math.Min(st, s);
                ed = Math.Max(ed, e);
            }
        }
        if (!insert) {
            ans.Add(new int[]{st, ed});
        }
        return ans.ToArray();
    }
}