看题目有点迷惑。但是连蒙带猜大概明白是啥意思了，结合数组的范围，推断出来是用BFS来找最小值。
一开始是用set来去重，很奇怪的是虽然报了超时，但是通过的用例是51/51......呵呵。但是至少证明是没猜错的，就是用BFS来搜索
public int minimumOperations(int[] nums, int start, int goal) {
Queue<Long> queue = new LinkedList<>();
queue.add((long) start);
int level = 0;
Set<Long> visited = new HashSet<>();
visited.add((long)start);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
long current = queue.poll();
if (current == goal) {
return level;
}
if (0 <= current && current <= 1000) {
for (int j = 0; j < nums.length; j++) {
if (!visited.contains(current + nums[j])) {
queue.add(current + nums[j]);
visited.add(current + nums[j]);
}
if (!visited.contains(current - nums[j])) {
queue.add(current - nums[j]);
visited.add(current - nums[j]);
}
if (!visited.contains(current ^ nums[j])) {
queue.add(current ^ nums[j]);
visited.add(current ^ nums[j]);
}
}
}
}
level++;
}
return -1;
}
很明显就是因为Set去重导致超时。于是转变想法用数组来表示是否重复，从而bfs剪枝。因此需要先明确数组的范围，看到题目。知道数值是限制在[0,1000]
于是直接这样子改造：
public int minimumOperations(int[] nums, int start, int goal) {
Queue<Integer> queue = new LinkedList<>();
queue.add(start);
int level = 0;
int[] visited = new int[1001];
visited[start] = 1;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
int current = queue.poll();
if (current == goal) {
return level;
}
if (0 <= current && current <= 1000) {
for (int j = 0; j < nums.length; j++) {
int a1 = current + nums[j];
int a2 = current - nums[j];
int a3 = current ^ nums[j];
if (visited[a1] == 0) {
queue.add(a1);
visited[a1] = 1;
}
if (visited[a2] == 0) {
queue.add(a2);
visited[a2] = 1;
}
if (visited[a3] == 0) {
queue.add(a3);
visited[a3] = 1;
}
}
}
}
level++;
}
return -1;
}
但是用测试用例发现下标会有-1这种负数导致越界。
于是又观察了下代码：
if (0 <= current && current <= 1000) {
for (int j = 0; j < nums.length; j++) {
int a1 = current + nums[j];
int a2 = current - nums[j];
int a3 = current ^ nums[j];
if (visited[a1] == 0) {
queue.add(a1);
visited[a1] = 1;
}
if (visited[a2] == 0) {
queue.add(a2);
visited[a2] = 1;
}
if (visited[a3] == 0) {
queue.add(a3);
visited[a3] = 1;
}
}
}
发现只有0 <= current && current <= 1000 才需要执行运算，那其实就可以把这个校验提前到这一轮的bfs，不用等到下一轮。于是改造成如下就pass了。
//比赛时写的代码，比较挫
public int minimumOperations(int[] nums, int start, int goal) {
Queue<Integer> queue = new LinkedList<>();
queue.add(start);
int level = 0;
int[] visited = new int[1010];
visited[start] = 1;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
int current = queue.poll();
if (current == goal) {
return level;
}
if (0 <= current && current <= 1000) {
for (int j = 0; j < nums.length; j++) {
int a1 = current + nums[j];
int a2 = current - nums[j];
int a3 = current ^ nums[j];
if (a1 == goal || a2 == goal || a3 == goal) {
return level + 1;
}
if (0 <= a1 && a1 <= 1000 && visited[a1] == 0) {
queue.add(a1);
visited[a1] = 1;
}
if (0 <= a2 && a2 <= 1000 && visited[a2] == 0) {
queue.add(a2);
visited[a2] = 1;
}
if (0 <= a3 && a3 <= 1000 && visited[a3] == 0) {
queue.add(a3);
visited[a3] = 1;
}
}
}
}
level++;
}
return -1;
}
