feat: 跳跃游戏 IIII

Author: sl1673495

BFS问题/跳跃游戏 IV-1345.js

@@ -0,0 +1,72 @@
+/**
+ * @param {number[]} arr
+ * @return {number}
+ */
+let minJumps = function (arr) {
+  let n = arr.length
+  if (n === 1) {
+    return 0
+  }
+
+  // 连续出现超过两次的数字就抛弃掉
+  let newArr = []
+  let sameCount = 0
+  for (let i = 0; i < arr.length; i++) {
+    if (arr[i] === arr[i - 1]) {
+      sameCount += 1
+      if (sameCount >= 2) {
+        continue
+      } else {
+        newArr.push(arr[i])
+      }
+    } else {
+      newArr.push(arr[i])
+      sameCount = 0
+    }
+  }
+  arr = newArr
+  n = arr.length
+  // 遍历一遍 记录每个数字出现的下标位置
+  let indexesMap = new Map()
+  for (let i = 0; i < n; i++) {
+    let val = arr[i]
+    let indexes = indexesMap.get(val)
+    if (!indexes) {
+      indexesMap.set(val, [i])
+    } else {
+      indexes.push(i)
+    }
+  }
+
+  let visited = []
+  let count = 0
+  let queue = [0]
+  while (queue.length) {
+    count++
+    let len = queue.length
+    for (let i = 0; i < len; i++) {
+      let index = queue.shift()
+      // 找到了 由于bfs的特性 此时用的跳跃次数一定是最少的
+      if (index === n - 1) {
+        return count - 1
+      }
+
+      // 没找到 继续把可以跳的几个位置都放入队列中
+      let val = arr[index]
+      let left = index - 1
+      let right = index + 1
+      let sameIndexes = indexesMap.get(val)
+
+      if (left >= 0 && !visited[left]) queue.push(left)
+      if (right < n && !visited[right]) queue.push(right)
+      for (let sameIndex of sameIndexes) {
+        if (sameIndex !== index && !visited[sameIndex]) {
+          queue.push(sameIndex)
+        }
+      }
+
+      visited[index] = true
+    }
+  }
+  return n
+}