[Swift]LeetCode45. 跳躍遊戲 II | Jump Game II

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Your goal is to reach the last index in the minimum number of jumps.

Example:

Input: [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2.
    Jump 1 step from index 0 to 1, then 3 steps to the last index.

Note:

You can assume that you can always reach the last index.

---

給定一個非負整數數組，你最初位於數組的第一個位置。

數組中的每一個元素表明你在該位置能夠跳躍的最大長度。

你的目標是使用最少的跳躍次數到達數組的最後一個位置。

示例:

輸入: [2,3,1,1,4]
輸出: 2
解釋: 跳到最後一個位置的最小跳躍數是 。
     從下標爲 0 跳到下標爲 1 的位置，跳  步，而後跳  步到達數組的最後一個位置。
213

說明:

假設你老是能夠到達數組的最後一個位置。

---

16ms

 1 class Solution {
 2     func jump(_ nums: [Int]) -> Int {
 3   if (nums.count == 0 || nums.count == 1) { return 0 }
 4   
 5   var res = 0
 6   var mi = 0
 7   for e in 0...nums.count-1 {
 8     if nums[e] == 0 {continue}
 9     var md = 0
10 
11     if e < mi {continue}
12     // print(e,mi)
13     for c in e+1...e + nums[e] {
14       if (c >= nums.count-1) { return res + 1}
15       if (c+nums[c] > md) {mi = c}
16       md = max(md,c+nums[c])
17       
18     }
19     res += 1
20   }
21   return res
22 }
23 }

---

80ms

 1 class Solution {
 2     func jump(_ nums: [Int]) -> Int {
 3 
 4         var startIndex = 0
 5         var endIndex = 0
 6         var jump = 0
 7         var mostFurther = 0
 8         for i in 0..<nums.count-1{
 9             mostFurther = max(mostFurther, i + nums[i])
10             
11             if i == endIndex{
12                 jump += 1
13                 endIndex = mostFurther
14             }
15             
16         }
17         return jump
18     }
19 }

---

96ms

 1 class Solution {
 2     func jump(_ nums: [Int]) -> Int {
 3         var cnt = 0, idx = 0
 4         var cur = 0, pre = 0
 5         while cur < nums.count - 1 {
 6             cnt += 1
 7             pre = cur
 8             while idx <= pre {
 9                 cur = max(cur, idx + nums[idx])
10                 idx += 1
11             }
12         }
13         return cnt
14     }
15 }