56.merge intervals

56.merge intervals

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 considerred overlapping.

// 4ms
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>

//Definition for an interval.
struct Interval {
    int start;
    int end;
};

int comp(const void *x, const void *y)
{
    struct Interval a = *(struct Interval *)x;
    struct Interval b = *(struct Interval *)y;
    // ASC
    return a.start - b.start;
}

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
struct Interval* merge(struct Interval* intervals, int intervalsSize, int* returnSize) {
    if(intervalsSize == 0) return intervals;

    struct Interval *result = (struct Interval *)malloc(sizeof(struct Interval) * intervalsSize);
    qsort(intervals, intervalsSize, sizeof(struct Interval), comp);

    *returnSize = 0;
    result[0].start = intervals[0].start;
    result[0].end = intervals[0].end;
    for (int i = 1; i < intervalsSize; ++i) {
        if(result[*returnSize].end < intervals[i].start)
        {
            ++*returnSize;
            result[*returnSize].start =  intervals[i].start;
            result[*returnSize].end =  intervals[i].end;
        }else if (result[*returnSize].end < intervals[i].end) {
            result[*returnSize].end =  intervals[i].end;
        }
        //(result[*returnSize].end >= intervals[i].end)
    }
    ++*returnSize;
    return result;
}

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