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[leedcode 45] Jump Game II

时间:2015-07-11 10:34:31      阅读:114      评论:0      收藏:0      [点我收藏+]

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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.)

public class Solution {
    public int jump(int[] nums) {
        //贪心思想,使用lastReach变量保存res-1步所能到达的最远路径,reach代表0~i所能到达的最远路径
        //当遍历的i大于lastReach时,res需要增加,同时更新lastReach为reach;
        int lastReach=0;
        int reach=0;
        int res=0;
        for(int i=0;i<nums.length;i++){
            if(lastReach<i){
                res++;
                lastReach=reach;
            }
            reach=Math.max(nums[i]+i,reach);
            
            
        }
        if(reach>=nums.length-1){
            return res;
        }else return -1;
    }
}

 

[leedcode 45] Jump Game II

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原文地址:http://www.cnblogs.com/qiaomu/p/4638049.html

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