Jump Game II

Jump Game II

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Question

Total Accepted: 52353 Total Submissions: 213392 Difficulty: Hard

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.)

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这个题目实际上用的是类似于动态规划的思想，借助于之前跳跃的次数可以到达的位置，然后计算在该位置上的跳跃次数。

public class Solution {    public int jump(int[] nums) {if(nums.length==0)return 0;        int [] result = new int [nums.length];        result[0]=0;        for(int i=1;i<nums.length;i++){        result[i]=Integer.MAX_VALUE;        }        for(int i=0;i<nums.length;i++){        for(int j=nums[i];j>=1;j--){//必须要从大到小遍历，才能保证一个值只计算一次        if(i+j<nums.length){        if(result[i+j]==Integer.MAX_VALUE){        result[i+j]=result[i]+1;        }        else{        break;        }        }        if(result[nums.length-1]!=Integer.MAX_VALUE){        return result[nums.length-1];        }        }        }        return result[nums.length-1];    }}