leetcode 1345. 跳跃游戏 IV
题目:1345. 跳跃游戏 IV - 力扣(LeetCode)
经典bfs,关键是建立所有“arr[i] == arr[j]”的连接。我的做法是用额外的存储,记录每个整数的前后整数都是哪个,再对数组排序。每个整数搜索的下个节点就是prev、next和数组中相邻且相等的整数:
struct Node {
int val;
int index;
int jumps = -1;
Node* prev = nullptr;
Node* next = nullptr;
Node(int val) {
this->val = val;
}
};
bool myComp(Node* a, Node* b) {
return a->val < b->val;
}
class Solution {
public:
int minJumps(vector<int>& arr) {
size_t n = arr.size();
if (n <= 1) {
return 0;
}
vector<Node*> nodes(n);
for (int i = 0; i < n; i++) {
nodes[i] = new Node(arr[i]);
if (i > 0) {
nodes[i - 1]->next = nodes[i];
nodes[i]->prev = nodes[i - 1];
}
}
list<Node*> bfs;
bfs.push_back(nodes[0]);
nodes[0]->jumps = 0;
Node* tail = nodes[n - 1];
sort(nodes.begin(), nodes.end(), myComp);
for (int i = 0; i < n; i++) {
nodes[i]->index = i;
}
Node* t;
int i;
while (!bfs.empty()) {
t = bfs.front();
bfs.pop_front();
i = t->index - 1;
while (i >= 0 && nodes[i]->val == t->val && nodes[i]->jumps == -1) {
nodes[i]->jumps = t->jumps + 1;
bfs.push_back(nodes[i]);
i--;
}
i = t->index + 1;
while (i < n && nodes[i]->val == t->val && nodes[i]->jumps == -1) {
nodes[i]->jumps = t->jumps + 1;
bfs.push_back(nodes[i]);
i++;
}
if (t->prev && t->prev->jumps == -1) {
t->prev->jumps = t->jumps + 1;
bfs.push_back(t->prev);
}
if (t->next && t->next->jumps == -1) {
t->next->jumps = t->jumps + 1;
bfs.push_back(t->next);
}
if (tail->jumps != -1) {
return tail->jumps;
}
}
return (int) n - 1;
}
};