[Greedy] B031_ merge interval (sort)

1. Title Description

Given a collection of intervals, merge all overlapping intervals.

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].

Second, the solution

Method 1: Greed

Wrong idea: I originally wanted to directly use end to record the value of the maximum end position in the first i-1 intervals of the i-th loop, and then use it to decide whether to merge the intervals:

• If they do not coincide, add the current array directly, followed by the new end value and endi subscript.
• Otherwise, let the right endpoint of the last element in the list be set to max(end, inte[i][0])

But there is a problem with this: when such a broken thing appears, it explodes ...[[1,4],[0,5]]

public int[][] merge(int[][] inte) {
    if (inte.length <= 1) {
        return inte;
    }
    List<int[]> list = new LinkedList<>();
    int end = inte[0][1], endi = 0;
    Arrays.sort(inte, (e1, e2) -> e1[0] - e2[0]);
    list.add(inte[0]);
    for (int i = 1; i < inte.length; i++) {
        if (end < inte[i][0]) { 	  //无交集
            list.add(inte[i]);
            end = inte[i][1];
            endi = i;
        } else if (end >= inte[i][0]){//交集
            list.get(endi)[1] = Math.max(inte[i][1], end);
        }
    }
    int[][] res = new int[list.size()][2];
    int i = 0;
    for (int[] arr : list) {
        res[i++] = arr;
    }
    return res;
}

Think about it, or judge the trouble on the linked list directly.

public int[][] merge(int[][] inte) {
    if (inte.length == 1) {
        return inte;
    }
    List<int[]> list = new ArrayList<>();
    Arrays.sort(inte, (e1, e2) -> e1[0] - e2[0]);

    for (int i = 0; i < inte.length; i++) {
        if (list.isEmpty() || list.get(list.size()-1)[1] < inte[i][0]) {
            list.add(inte[i]);
        } else {
            int end = list.get(list.size()-1)[1];
            list.get(list.size()-1)[1] = Math.max(end, inte[i][1]);
        }
    }
    int[][] res = new int[list.size()][2];
    int i = 0;
    for (int[] arr : list) {
        res[i++] = arr;
    }
    return res;
}

Complexity analysis

• time complexity: $O()$，
• Space complexity: $O()$，