标签：oar 图片 技术 return new ISE size desire ril
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:
0 <= A.length < 10000 <= B.length < 10000 <= A[i].start, A[i].end, B[i].start, B[i].end < 10^9NOTE: 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
986. Interval List Intersections
标签：oar 图片 技术 return new ISE size desire ril
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/12945392.html
