Interval Interval scheduling problem of intersection

Interval Interval scheduling problem of intersection

Interval Scheduling wrote a total of 3 blog, the first two overlapping sections and sections merger were told interval 最大不相交子集和重叠区间的合并, write an algorithm today, you can 快速找出两组区间的交集.

First, the problem-solving ideas

Interval ideas to solve the problem in general is 先排序, in order to operate, but the topic that has been sorted, then you can 用两个索引指针在 A 和 B 中游走，把交集找出来code something like this:

# A, B 形如 [[0,2],[5,10]...]
def intervalIntersection(A, B):
    i, j = 0, 0
    res = []
    while i < len(A) and j < len(B):
        # ...
        j += 1
        i += 1
    return res

Difficult, let's honestly analyze various situations.

First, for two intervals, we use [a1, a2] and [b1, b2] represents the two sections A and B, then there is no intersection of these two intervals under what circumstances it :

Only these two cases, written conditional code is this:

if b2 < a1 or a2 < b1:
    [a1,a2] 和 [b1,b2] 无交集

So, under what circumstances, there is the intersection of two intervals it? According to deny the proposition, the logic of the above proposition is whether the presence of intersection:

# 不等号取反，or 也要变成 and
if b2 >= a1 and a2 >= b1:
    [a1,a2] 和 [b1,b2] 存在交集

Next, the case of two intervals intersection exists, what does? 穷举come out:

It is very simple, it is these four cases only. So then I think, under these types of situations, whether there is an intersection in common?

We were surprised to find that the intersection interval is regular! If the intersection is the interval [c1, c2],Then c1 = max (a1, b1), c2 = min (a2, b2)! This is to find the intersection of the core , we further the code:

while i < len(A) and j < len(B):
    a1, a2 = A[i][0], A[i][1]
    b1, b2 = B[j][0], B[j][1]
    if b2 >= a1 and a2 >= b1:
        res.append([max(a1, b1), min(a2, b2)])
    # ...

The final step, we pointer i and j are sure to advance (increment), and when should it go forward?

Whether 只取决于 a2 和 b2 的大小关系forward . Code:

while i < len(A) and j < len(B):
    # ...
    if b2 < a2:
        j += 1
    else:
        i += 1

Complete code:

# A, B 形如 [[0,2],[5,10]...]
def intervalIntersection(A, B):
    i, j = 0, 0 # 双指针
    res = []
    while i < len(A) and j < len(B):
        a1, a2 = A[i][0], A[i][1]
        b1, b2 = B[j][0], B[j][1]
        # 两个区间存在交集
        if b2 >= a1 and a2 >= b1:
            # 计算出交集，加入 res
            res.append([max(a1, b1), min(a2, b2)])
        # 指针前进
        if b2 < a2: j += 1
        else:       i += 1
    return res

C ++ code:

class Solution {
public:
    vector<vector<int>> intervalIntersection(vector<vector<int>>& A, vector<vector<int>>& B) {
        //保存结果
        vector<vector<int>> ret;
        if(A.empty() || B.empty())
        {
            return ret;
        }

        //i和j两个下标索引
        int i = 0;
        int j = 0;
        while(i < A.size() && j < B.size())
        {
            //下面四个变量的含义相见博客
            int a1 = A[i][0];
            int a2 = A[i][1];
            int b1 = B[j][0];
            int b2 = B[j][1];

            //代表区间有交集，自己画个图就OK
            if(b2 >= a1 && a2 >= b1)
            {
                //max(a1,b1),min(a2,b2)两个区间的交集部分
                ret.push_back({max(a1,b1),min(a2,b2)});
            }

            //更新i和j下标索引，b2 < a2 就代表i所在的区间大于j所在的区间，更新j
            //是因为i区间长于j区间的部分还可能和j的下一个区间继续重叠，还需要继续判断
            //所以更新j而不是更新i
            if(b2 < a2)
            {
                j++;
            }
            else//反之同理
            {
                i++;
            }
        }   
        return ret;
    }
};

To summarize, the section looks like the problem is more complex, difficult to handle a lot of situations, but in fact can be found by observing the law of commonality between different situations, can handle with simple code.