[LeetCode] Jump Game II

原创于 2014-12-26 10:04:53 发布 · 553 阅读

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本文介绍了一个使用贪心法解决的算法问题：给定一个非负整数数组，每个元素代表当前位置能够跳跃的最大长度，目标是从数组起始位置到达末尾位置所需的最少跳跃次数。文章通过示例展示了如何实现这一算法。

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.)

思路：贪心法。

代码：

class Solution {
public:
    int jump(int A[], int n) {  
        int maxx=0,temp=0,num=0;  
        for(int i=0;i<n;)  
        {  
            if(temp>=n-1)break;  
            while(i<=temp)  
            {  
                maxx=max(maxx,i+A[i]);  
                ++i;  
            }  
            num++;  
            temp=maxx;              
        }  
        return num;  
          
    } 
};