【每日算法】一题五解：找「两条链表的第一个公共节点」｜Python 主题月

输入：intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
输出：Reference of the node with value = 8
输入解释：相交节点的值为 8 （注意，如果两个列表相交则不能为 0）。从各自的表头开始算起，链表 A 为 [4,1,8,4,5]，链表 B 为 [5,0,1,8,4,5]。在 A 中，相交节点前有 2 个节点；在 B 中，相交节点前有 3 个节点。
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输入：intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
输出：Reference of the node with value = 2
输入解释：相交节点的值为 2 （注意，如果两个列表相交则不能为 0）。从各自的表头开始算起，链表 A 为 [0,9,1,2,4]，链表 B 为 [3,2,4]。在 A 中，相交节点前有 3 个节点；在 B 中，相交节点前有 1 个节点。
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输入：intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
输出：null
输入解释：从各自的表头开始算起，链表 A 为 [2,6,4]，链表 B 为 [1,5]。由于这两个链表不相交，所以 intersectVal 必须为 0，而 skipA 和 skipB 可以是任意值。
解释：这两个链表不相交，因此返回 null。
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public class Solution {
    public ListNode getIntersectionNode(ListNode a, ListNode b) {
        for (ListNode h1 = a; h1 != null ; h1 = h1.next) {
            for (ListNode h2 = b; h2 != null ; h2 = h2.next) {
                if (h1 == h2) return h1;
            }
        }
        return null;
    }
}
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class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        h1 = headA
        while h1:
            h2 = headB
            while h2:
                if h1 == h2:
                    return h1
                h2 = h2.next
            h1 = h1.next
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public class Solution {
    public ListNode getIntersectionNode(ListNode a, ListNode b) {
        Deque<ListNode> d1 = new ArrayDeque<>(), d2 = new ArrayDeque<>();
        while (a != null) {
            d1.add(a);
            a = a.next;
        }
        while (b != null) {
            d2.add(b);
            b = b.next;
        }
        ListNode ans = null;
        while (!d1.isEmpty() && !d2.isEmpty() && d1.peekLast() == d2.peekLast()) {
            ans = d1.pollLast();
            d2.pollLast();
        }
        return ans;
    }
}
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class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        d1, d2 = deque([]), deque([])
        while headA:
            d1.append(headA)
            headA = headA.next
        while headB:
            d2.append(headB)
            headB = headB.next
        ans = None
        while d1 and d2 and d1[-1] == d2[-1]:
            ans = d1.pop()
            d2.pop()
        return ans
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public class Solution {
    public ListNode getIntersectionNode(ListNode a, ListNode b) {
        Set<ListNode> set = new HashSet<>();
        while (a != null) {
            set.add(a);
            a = a.next;
        }
        while (b != null && !set.contains(b)) b = b.next;
        return b;
    }
}
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class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        hashSet = set()
        while headA:
            hashSet.add(headA)
            headA = headA.next
        while headB and headB not in hashSet:
            headB = headB.next
        return headB
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public class Solution {
    public ListNode getIntersectionNode(ListNode a, ListNode b) {
        int c1 = 0, c2 = 0;
        ListNode ta = a, tb = b;
        while (ta != null && c1++ >= 0) ta = ta.next;
        while (tb != null && c2++ >= 0) tb = tb.next;
        int d = c1 - c2;
        if (d > 0) {
            while (d-- > 0) a = a.next;
        } else if (d < 0) {
            d = -d;
            while (d-- > 0) b = b.next;
        }
        while (a != b) {
            a = a.next;
            b = b.next;
        }
        return a;
    }
}
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class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        c1 = c2 = 0
        ta, tb = headA, headB
        while ta:
            ta = ta.next
            c1 += 1
        while tb:
            tb = tb.next
            c2 += 1
        d = c1 - c2
        if d > 0:
            while d:
                headA = headA.next
                d -= 1
        elif d < 0:
            d = -d
            while d:
                headB = headB.next
                d -= 1
        while headA != headB:
            headA = headA.next
            headB = headB.next
        return headA
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public class Solution {
    public ListNode getIntersectionNode(ListNode a, ListNode b) {
        ListNode ta = a, tb = b;
        while (ta != tb) {
            ta = ta == null ? b : ta.next;
            tb = tb == null ? a : tb.next;
        }
        return ta;
    }
}
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class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        ta, tb = headA, headB
        while ta != tb:
            ta = headB if not ta else ta.next
            tb = headA if not tb else tb.next
        return ta
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