986. Interval List Intersections

Given two lists of closed intervals, each list of intervals is pairwise disjoint and in sorted order.

Return the intersection of these two interval lists.

(Formally, a closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b.  The intersection of two closed intervals is a set of real numbers that is either empty, or can be represented as a closed interval.  For example, the intersection of [1, 3] and [2, 4] is [2, 3].)

Example 1:

Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
Reminder: The inputs and the desired output are lists of Interval objects, and not arrays or lists.

Note:

1. 0 <= A.length < 1000
2. 0 <= B.length < 1000
3. 0 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9

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

class Solution {
    public int[][] intervalIntersection(int[][] A, int[][] B) {
        Arrays.sort(A, (a, b)->a[0]-b[0]);//可以不要
        Arrays.sort(B, (a, b)->a[0]-b[0]);
        if(A.length == 0 || B.length == 0) return new int[][]{};
        List<int[]> list = new ArrayList();
        for(int i = 0; i < A.length; i++){
            int[] t1 = A[i];
            for(int j = 0; j < B.length; j++){
                int[] t2 = B[j];
                if(t1[1] < t2[0] || t1[0] > t2[1]) continue;
                int[] k = new int[2];
                k[0] = Math.max(t1[0], t2[0]);
                k[1] = Math.min(t1[1], t2[1]);
                list.add(k);
            }
        }
        int[][] result = new int[list.size()][2];
        for(int i = 0; i < list.size(); i++) result[i] = list.get(i);
        return result;
    }
}

似曾相识啊，跟merge intervals挺像的，先跳过不可能的情况，然后把合适的插入即可

class Solution {
  public int[][] intervalIntersection(int[][] A, int[][] B) {
    List<int[]> ans = new ArrayList();
    int i = 0, j = 0;

    while (i < A.length && j < B.length) {
      // Let's check if A[i] intersects B[j].
      // lo - the startpoint of the intersection
      // hi - the endpoint of the intersection
      int lo = Math.max(A[i][0], B[j][0]);
      int hi = Math.min(A[i][1], B[j][1]);
      if (lo <= hi)
        ans.add(new int[]{lo, hi});

      // Remove the interval with the smallest endpoint
      if (A[i][1] < B[j][1])
        i++;
      else
        j++;
    }

    return ans.toArray(new int[ans.size()][]);
  }
}

优化之后