【贪心】45. Jump Game II

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原创最新推荐文章于 2026-03-17 02:03:24 发布 170 阅读 AI 写同款· GEO 优化 ›

LeetCode 45. Jump Game II--当前元素值表示能跳跃的最大长度，求跳跃到最后一个元素的最少跳跃次数--贪心算法 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... 阅读详情

与55. JUMP GAME 思路相同，加上一个统计步幅次数的变量即可：

class Solution {
public:
    int jump(vector<int>& nums) {
        if(nums.size()<2)
            return 0;
        
        int i=0;  
        int res=1;
        while(i<nums.size())  
        {  
            int max_index=i+nums[i],index=i;         
              
            for(int j=i;j<=i+nums[i];j++)  
            {  
                if(j+nums[j]>=nums.size()-1) 
                    return j>i?res+1:res;  
              
                if( max_index <= (j+nums[j]) )  
                {  
                    max_index= j + nums[j];  
                    index=j;                   
                }  
                  
            }  
              
            i=index;  
            res++;
        }  
    }
};

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