Question

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

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

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()][]);
  }
}

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;
  }
};

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
}

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;
}

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
  }
}