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LeetCode #986 Interval List Inte

LeetCode #986 Interval List Inte

作者: air_melt | 来源:发表于2022-01-12 21:30 被阅读0次

    986 Interval List Intersections 区间列表的交集

    Description:
    You are given two lists of closed intervals, firstList and secondList, where firstList[i] = [starti, endi] and secondList[j] = [startj, endj]. Each list of intervals is pairwise disjoint and in sorted order.

    Return the intersection of these two interval lists.

    A closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b.

    The intersection of two closed intervals is a set of real numbers that are either empty or represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].

    Example:

    Example 1:

    interval

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

    Example 2:

    Input: firstList = [[1,3],[5,9]], secondList = []
    Output: []

    Constraints:

    0 <= firstList.length, secondList.length <= 1000
    firstList.length + secondList.length >= 1
    0 <= starti < endi <= 10^9
    endi < starti+1
    0 <= startj < endj <= 10^9
    endj < startj+1

    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 :

    示例 1:

    区间列表的交集

    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 <= 10^9
    endi < starti+1
    0 <= startj < endj <= 10^9
    endj < startj+1

    思路:

    双指针
    区间已经有序所以交集左边界一定是 max(firstList[i][0], secondList[j][0]), 右边界一定是 min(firstList[i][1], secondList[j][1])
    如果相交 right >= left
    否则不相交, 这时需要移动指针, 比较 firstList[i][1], secondList[j][1], 较小的指针往后移
    时间复杂度为 O(n), 空间复杂度为 O(1)

    代码:
    C++:

    class Solution 
    {
    public:
        vector<vector<int>> intervalIntersection(vector<vector<int>>& firstList, vector<vector<int>>& secondList) 
        {
            int i = 0, j = 0, m = firstList.size(), n = secondList.size();
            vector<vector<int>> result;
            while (i < m and j < n) 
            {
                int left = max(firstList[i].front(), secondList[j].front()), right = min(firstList[i].back(), secondList[j].back());
                if (right >= left) result.emplace_back(vector<int>{left, right});
                if (firstList[i].back() < secondList[j].back()) ++i;
                else ++j;
            }
            return result;
        }
    };
    

    Java:

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

    Python:

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

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