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986. 区间列表的交集

English Version

题目描述

给定两个由一些 闭区间 组成的列表，firstList 和 secondList ，其中 firstList[i] = [starti, endi] 而 secondList[j] = [startj, endj] 。每个区间列表都是成对 不相交 的，并且 已经排序 。

返回这 两个区间列表的交集 。

形式上，闭区间 [a, b]（其中 a <= b）表示实数 x 的集合，而 a <= x <= b 。

两个闭区间的 交集 是一组实数，要么为空集，要么为闭区间。例如，[1, 3] 和 [2, 4] 的交集为 [2, 3] 。

 

示例 1：

输入：firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]]
输出：[[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]

示例 2：

输入：firstList = [[1,3],[5,9]], secondList = []
输出：[]

示例 3：

输入：firstList = [], secondList = [[4,8],[10,12]]
输出：[]

示例 4：

输入：firstList = [[1,7]], secondList = [[3,10]]
输出：[[3,7]]

 

提示：

  • 0 <= firstList.length, secondList.length <= 1000
  • firstList.length + secondList.length >= 1
  • 0 <= starti < endi <= 109
  • endi < starti+1
  • 0 <= startj < endj <= 109
  • endj < startj+1

解法

方法一：双指针

Python3

class Solution:
    def intervalIntersection(
        self, firstList: List[List[int]], secondList: List[List[int]]
    ) -> List[List[int]]:
        i = j = 0
        ans = []
        while i < len(firstList) and j < len(secondList):
            s1, e1, s2, e2 = *firstList[i], *secondList[j]
            l, r = max(s1, s2), min(e1, e2)
            if l <= r:
                ans.append([l, r])
            if e1 < e2:
                i += 1
            else:
                j += 1
        return ans

Java

class Solution {
    public int[][] intervalIntersection(int[][] firstList, int[][] secondList) {
        List<int[]> ans = new ArrayList<>();
        int m = firstList.length, n = secondList.length;
        for (int i = 0, j = 0; i < m && j < n;) {
            int l = Math.max(firstList[i][0], secondList[j][0]);
            int r = Math.min(firstList[i][1], secondList[j][1]);
            if (l <= r) {
                ans.add(new int[] {l, r});
            }
            if (firstList[i][1] < secondList[j][1]) {
                ++i;
            } else {
                ++j;
            }
        }
        return ans.toArray(new int[ans.size()][]);
    }
}

C++

class Solution {
public:
    vector<vector<int>> intervalIntersection(vector<vector<int>>& firstList, vector<vector<int>>& secondList) {
        vector<vector<int>> ans;
        int m = firstList.size(), n = secondList.size();
        for (int i = 0, j = 0; i < m && j < n;) {
            int l = max(firstList[i][0], secondList[j][0]);
            int r = min(firstList[i][1], secondList[j][1]);
            if (l <= r) ans.push_back({l, r});
            if (firstList[i][1] < secondList[j][1])
                ++i;
            else
                ++j;
        }
        return ans;
    }
};

Go

func intervalIntersection(firstList [][]int, secondList [][]int) [][]int {
	m, n := len(firstList), len(secondList)
	var ans [][]int
	for i, j := 0, 0; i < m && j < n; {
		l := max(firstList[i][0], secondList[j][0])
		r := min(firstList[i][1], secondList[j][1])
		if l <= r {
			ans = append(ans, []int{l, r})
		}
		if firstList[i][1] < secondList[j][1] {
			i++
		} else {
			j++
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

TypeScript

function intervalIntersection(
    firstList: number[][],
    secondList: number[][],
): number[][] {
    const n = firstList.length;
    const m = secondList.length;
    const res = [];
    let i = 0;
    let j = 0;
    while (i < n && j < m) {
        const start = Math.max(firstList[i][0], secondList[j][0]);
        const end = Math.min(firstList[i][1], secondList[j][1]);
        if (start <= end) {
            res.push([start, end]);
        }
        if (firstList[i][1] < secondList[j][1]) {
            i++;
        } else {
            j++;
        }
    }
    return res;
}

Rust

impl Solution {
    pub fn interval_intersection(
        first_list: Vec<Vec<i32>>,
        second_list: Vec<Vec<i32>>,
    ) -> Vec<Vec<i32>> {
        let n = first_list.len();
        let m = second_list.len();
        let mut res = Vec::new();
        let (mut i, mut j) = (0, 0);
        while i < n && j < m {
            let start = first_list[i][0].max(second_list[j][0]);
            let end = first_list[i][1].min(second_list[j][1]);
            if start <= end {
                res.push(vec![start, end]);
            }
            if first_list[i][1] < second_list[j][1] {
                i += 1;
            } else {
                j += 1;
            }
        }
        res
    }
}

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