118、合并区间

题目描述：
给出一个区间的集合，请合并所有重叠的区间。

示例 1:

输入: [[1,3],[2,6],[8,10],[15,18]]
输出: [[1,6],[8,10],[15,18]]
解释: 区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].
示例 2:

输入: [[1,4],[4,5]]
输出: [[1,5]]
解释: 区间 [1,4] 和 [4,5] 可被视为重叠区间。

代码：

class Solution {
    public int[][] merge(int[][] intervals) {
    
         if(intervals.length == 0){
		   return intervals;
	   }
        PriorityQueue<int[]> tempri = new PriorityQueue<>(
			   (a,b)->(a[0] - b[0])
			   );
	   
	   for (int[] is : intervals) {
		tempri.offer(is);
	   		}
	   Stack<int[]> temStack = new Stack<>();
	   temStack.push(tempri.peek());
	   while (!tempri.isEmpty()) {
		 int tem[] =  tempri.poll();
		  int stack[] = temStack.pop();
		  if(tem[0] >= stack[0] && tem[1] <= stack[1]){
			  temStack.push(stack);
			  continue;
		  }
		  
		  
		  if (tem[0] <= stack[1] && stack[1] <= tem[1]) {
			  stack[1] = tem[1];
			  temStack.push(new int[]{stack[0],stack[1]});
		  }else {
			  temStack.push(new int[]{stack[0],stack[1]});
			  temStack.push(new int[]{tem[0],tem[1]});
		}
	   }
	  int[][]result = new int[temStack.size()][2];

	   int index = 0;
	   for (int[] is : temStack) {
			result[index][0] = is[0];
			result[index++][1] = is[1];
		}    
	   return result;
    }
}

排名靠前的代码

class Solution {
    public int[][] merge(int[][] intervals) {
        int rows = intervals.length;
        if (rows == 0) {
            return intervals;
        }
        int count = rows;
        for (int i = 0; i < rows - 1; i++) {
            for (int j = i + 1; j < rows; j++) {
                if (canMerge(intervals, i, j)) {
                    count--;
                    break;
                }
            }
        }
        int[][] res = new int[count][2];
        int index = 0;
        for (int[] row : intervals) {
            if (row != null) {
                res[index++] = row;
            }
        }
        return res;
    }

    public boolean canMerge(int[][] intervals, int first, int second) {
        if (!((intervals[first][0] < intervals[second][0] && intervals[first][1] < intervals[second][0])
                || (intervals[second][0] < intervals[first][0] && intervals[second][1] < intervals[first][0]))) {
            intervals[second] = new int[] { Math.min(intervals[first][0], intervals[second][0]),
                    Math.max(intervals[first][1], intervals[second][1]) };
            intervals[first] = null;
            return true;
        }
        return false;
    }
}