Skip to content

Latest commit

 

History

History
178 lines (140 loc) · 4.94 KB

File metadata and controls

178 lines (140 loc) · 4.94 KB

English Version

题目描述

给定两个由一些 闭区间 组成的列表,firstListsecondList ,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][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

双指针实现区间合并。

Python3

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

Java

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

C++

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

Go

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

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
}

...