lc2674

快慢指针

while (f->next != list && f->next->next != list)

得到的slow节点 中/偏前

ListNode* second = s->next;

// 后半 头节点

class Solution {
public:
vector<ListNode*> splitCircularLinkedList(ListNode* list)
{
vector<ListNode*> ret;
ListNode* s = list;
ListNode* f = list;
//f->next != list && f->next->next != list
while (f->next != list && f->next->next != list)
{
f = f->next->next;
s = s->next;
}
ListNode* second = s->next;
// 后半 头节点

s->next = list; // 前半 循环

ListNode* tail = second;
while (tail->next != list)
tail = tail->next;
tail->next = second;//后半 循环

return {list, second};
}
};

lc1428

逐行对每行链表用二分找首个1的列号

取所有行结果的最小列号，无1则返回-1。

class Solution {

public:

int leftMostColumnWithOne(BinaryMatrix &m) {

auto d = m.dimensions();

int r = d[0], c = d[1], res = -1;

auto bs = [&](int i, int r) {

int l = 0, k = -1;

while (l <= r)

{

int mid = l + ((r - l) >> 1);

if (m.get(i, mid) == 0)

l = mid + 1;

else

{

k = mid;

r = mid - 1;

}

}

return k;

};

for (int i = 0; i < r; i++) {

int k = bs(i, c - 1);

if (k >= 0) {

if (res == -1)

res = k;

else

res = min(res, k);

}

}

return res;

}

};