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| class Solution {
public int minJumps(int[] arr) {
Map<Integer, Set<Integer>> numMap = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
numMap.computeIfAbsent(arr[i], a -> new HashSet<>()).add(i);
}
Deque<Integer> q = new ArrayDeque<>();
boolean[] visited = new boolean[arr.length];
q.offer(0);
visited[0] = true;
int steps = 0;
while (!q.isEmpty()) {
for (int i = q.size(); i > 0; i--) {
int currPos = q.poll();
if (currPos == arr.length - 1) {
return steps;
}
int left = currPos - 1;
if (left >= 0 && !visited[left]) {
q.offer(left);
visited[left] = true;
}
int right = currPos + 1;
if (right < arr.length && !visited[right]) {
q.offer(right);
visited[right] = true;
}
if (numMap.containsKey(arr[currPos])) {
Set<Integer> neighbors = numMap.get(arr[currPos]);
for (Integer neighbor : neighbors) {
if (!visited[neighbor]) {
q.offer(neighbor);
visited[neighbor] = true;
}
}
numMap.remove(arr[currPos]);
}
}
steps += 1;
}
return -1;
}
}
|
BFS就完事儿了. 注意不能在遍历hashset的时候删除元素. 第36行很重要, 避免重复遍历neighbors. 比如index 1, 3, 7都是元素5, 我们先到达1, 然后把3和7压入queue中, 当我们到3时, 就不要再遍历它的neighbors了, 因为它的neighors在index是1的时候就都添加进queue中了. 这个很重要否则会TLE.
时间复杂度: O(n)
空间复杂度: O(n)
