【leetcode刷題筆記】Jump Game II

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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.

For example:
Given array A = [2,3,1,1,4]

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.)

 

題解：跟Jump Game十分相像，把原來的boolean數組改成int型數組，用來記錄到達i時候最小的步驟數目，遍曆j(j<i)，考察從j跳到i需要的步驟數目是否小於當前的reach[i]，如果小於，就更新reach[i]。最終返回reach[A.length-1]即可。

代碼如下：

 1 public class Solution { 2     public int jump(int[] A) { 3         if(A == null || A.length == 0) 4             return 0; 5          6         int[] reach = new int[A.length]; 7         Arrays.fill(reach, Integer.MAX_VALUE); 8         reach[0] = 0; 9         10         for(int i = 1;i < A.length;i++){11             for(int j = 0;j < i;j++){12                 if(reach[j] != Integer.MAX_VALUE && j+A[j]>=i){13                     reach[i] = Math.min(reach[i], reach[j]+1);14                     break;15                 }16             }17         }18         19         return reach[A.length-1];20     }21 }