[LeetCode] 56. Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

合并区间。

题意是给一个二维数组，其中每个元素给的是一个类似[start, end]的范围。要求把这些区间尽可能地merge在一起。这题又是用到扫描线的思想。首先需要根据start对input排序，然后从第二个interval SECOND开始，跟其之前的interval FIRST作比较。比较的方式是

• 如果SECOND[start] > FIRST[end] - 说明两个interval没有交集，直接把FIRST加入最后的结果集
• 如果SECOND[start] <= FIRST[end] - 说明两个interval有交集，需要merge。merge后的interval的start自然是FIRST[start]，但是merge后的end是比较FIRST[end]和SECOND[end]谁大

时间O(nlogn) - 因为有sort

空间O(n) - 存放了最后的输出

JavaScript实现

/**
 * @param {number[][]} intervals
 * @return {number[][]}
 */
var merge = function (intervals) {
    // corner case
    if (intervals === null || intervals.length === 0) {
        return intervals;
    }

    // normal case
    let res = [];
    intervals.sort((a, b) => a[0] - b[0]);
    let cur = intervals[0];
    res.push(cur);
    for (let interval of intervals) {
        if (interval[0] <= cur[1]) {
            cur[1] = Math.max(cur[1], interval[1]);
        } else {
            cur = interval;
            res.push(cur);
        }
    }
    return res;
};

Java实现

class Solution {
    public int[][] merge(int[][] intervals) {
        // corner case
        if (intervals.length <= 1) {
            return intervals;
        }

        // normal case
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        List<int[]> res = new ArrayList<>();
        int[] cur = intervals[0];
        res.add(cur);
        for (int[] interval : intervals) {
            if (interval[0] <= cur[1]) {
                cur[1] = Math.max(cur[1], interval[1]);
            } else {
                cur = interval;
                res.add(cur);
            }
        }
        return res.toArray(new int[res.size()][]);
    }
}

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