56. Merge Intervals (JAVA)

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.

涉及：

1. 数组的截取与拷贝
2. 重写Array.sort的compare函数（注意：jdk1.7以后，必须要定义相等的情况返回0，否则会报Runtime error)

class Solution {
    public int[][] merge(int[][] intervals) {
        if(intervals.length == 0) return intervals;
        
        Arrays.sort(intervals,0, intervals.length, new ArrayComparator());

        int curIndex = 0;
        for(int i = 1; i < intervals.length; i++){
            if(intervals[curIndex][1] >= intervals[i][1]){ //only reserve intervals[i-1]
                continue;
            }
            else if(intervals[curIndex][1] >= intervals[i][0]){//integrate
                intervals[curIndex][1] = intervals[i][1];
            }
            else{ //no interval
                curIndex++;
                intervals[curIndex][0] = intervals[i][0];
                intervals[curIndex][1] = intervals[i][1];
            }
        }
        curIndex++;
        int[][] ret = new int[curIndex][2];
        System.arraycopy(intervals, 0, ret, 0, curIndex);
        
        return ret;
    }
}

class ArrayComparator implements Comparator{
    public int compare(Object o1, Object o2){
        int[] a = (int[]) o1;
        int[] b = (int[]) o2;
        if(a[0] > b[0]) return 1;//ascending order
        else if(a[0] < b[0]) return -1;
        else return 0;
    }
}