How to Compute the Interval List Intersections using Two Pointer

  • 时间：2020-09-16 12:48:17
  • 分类：网络文摘
  • 阅读：117 次

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:

merge-intervals How to Compute the Interval List Intersections using Two Pointer Algorithms? algorithms c / c++

merge-intervals

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 < 1000
0 <= B.length < 1000
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.

Merge Intervals using Two Pointers

Since the intervals are sorted, we can have two pointers iteratedly incrementing at two sides. There are 3 pairs (6 circumstances) regarding the relations between interval A and B. We then can deal with them accordingly.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

class Solution {
public:
    vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
        vector<vector<int>> r;
        int n = A.size();
        int m = B.size();
        int i = 0, j = 0;
        while ((i < n) && (j < m)) {
            if (A[i][1] < B[j][0]) {
                i ++;
                continue;
            }
            if (B[j][1] < A[i][0]) {
                j ++;
                continue;
            }
            if ((A[i][0] >= B[j][0]) && (A[i][1] <= B[j][1])) {
                r.push_back(A[i]);
                i ++;
                continue;
            }
            if ((B[j][0] >= A[i][0]) && (B[j][1] <= A[i][1])) {
                r.push_back(B[j]);
                j ++;
                continue;
            }                
            if ((A[i][0] <= B[j][0]) && (A[i][1] <= B[j][1])) {
                r.push_back({B[j][0], A[i][1]});
                i ++;
                continue;
            }
            if ((B[j][0] <= A[i][0]) && (B[j][1] <= A[i][1])) {
                r.push_back({A[i][0], B[j][1]});
                j ++;
                continue;
            }            
        }
        return r;
    }
};
class Solution {
public:
    vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
        vector<vector<int>> r;
        int n = A.size();
        int m = B.size();
        int i = 0, j = 0;
        while ((i < n) && (j < m)) {
            if (A[i][1] < B[j][0]) {
                i ++;
                continue;
            }
            if (B[j][1] < A[i][0]) {
                j ++;
                continue;
            }
            if ((A[i][0] >= B[j][0]) && (A[i][1] <= B[j][1])) {
                r.push_back(A[i]);
                i ++;
                continue;
            }
            if ((B[j][0] >= A[i][0]) && (B[j][1] <= A[i][1])) {
                r.push_back(B[j]);
                j ++;
                continue;
            }                
            if ((A[i][0] <= B[j][0]) && (A[i][1] <= B[j][1])) {
                r.push_back({B[j][0], A[i][1]});
                i ++;
                continue;
            }
            if ((B[j][0] <= A[i][0]) && (B[j][1] <= A[i][1])) {
                r.push_back({A[i][0], B[j][1]});
                j ++;
                continue;
            }            
        }
        return r;
    }
};

A and B are either: disjoints, inclusive or interleaving.

It could be improved into the following much more concise solution where we calculate the max of the lower bounds and the min of the higher bounds. Then, the interval is pushed if it is valid.

Then we increment the pointer of the interval where it has a smaller upperbound. 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

class Solution {
public:
    vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
        vector<vector<int>> r;
        int n = A.size();
        int m = B.size();
        int i = 0, j = 0;
        while ((i < n) && (j < m)) {
            int start = max(A[i][0], B[j][0]);
            int end = min(A[i][1], B[j][1]);
            if (start <= end) {
                r.push_back({start, end});
            }
            if (A[i][1] < B[j][1]) {
                i ++;
            } else {
                j ++;
            }
        }
        return r;
    }
};
class Solution {
public:
    vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
        vector<vector<int>> r;
        int n = A.size();
        int m = B.size();
        int i = 0, j = 0;
        while ((i < n) && (j < m)) {
            int start = max(A[i][0], B[j][0]);
            int end = min(A[i][1], B[j][1]);
            if (start <= end) {
                r.push_back({start, end});
            }
            if (A[i][1] < B[j][1]) {
                i ++;
            } else {
                j ++;
            }
        }
        return r;
    }
};

Both solutions run in O(M + N) time, and require O(1) constant space (excluding the result vector).

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
三种食物滋阴润燥蜂蜜是冬天补养佳品　 川贝雪梨适合秋咳 甘草片止咳不治咳　 花生营养丰富怎么吃对人体健康最有益　 吃水果的禁忌以及富含维生素的水果　 黄瓜的营养价值保健效果及食用方法　 土鸡蛋与洋鸡蛋的营养价值以及区别　 冬季适当吃糯米类食物有御寒滋补之功效　 三类护耳食物可以延缓老年人听力下降　 红枣养生知识：食用红枣需注意的问题　 健康面点拒绝这些有毒的添加剂原料　
评论列表
添加评论