LeetCode刷题笔记 LCP 09. 最小跳跃次数

BFS

bfs 不超时 有例子易懂

class Solution {
public:
    int minJump(vector<int>& jump) {
        int N=jump.size();
        vector<int> vis(N,0);
        queue<int> q;
        queue<int> num;
        //把0~N-1放入num
        for(int i=0;i<N;i++)
            num.push(i);
        //从0开始
        q.push(0);
        vis[0]=1;
        
        int dist=0;
        //bfs
        while(!q.empty()){
            int size=q.size();
            for(int i=0;i<size;i++){
                int tmp=q.front();
                q.pop();
                //是否结束
                if(tmp+jump[tmp]>=N)
                    return dist+1;
                //向右找
                if(vis[tmp+jump[tmp]]==0){
                    vis[tmp+jump[tmp]]=1;
                    q.push(tmp+jump[tmp]);
                }
                //向左找
                while(!num.empty() && num.front()<tmp){ / 向左弹射到任意左侧弹簧的位置
                    int top=num.front();
                    num.pop();
                    if(vis[top]==0){
                        vis[top]=1;
                        q.push(top);
                    }
                }
            }
            ++dist;
        }
        return -1;
    }
};

愣是没看懂，留着等个缘分

class Solution {
    int d[1000005],q[1000005],he,ta;
public:
    int minJump(vector<int>& jump) {
        int n=jump.size(),i,l=0;
        fill(d,d+n,-1);
        d[0]=0;
        he=ta=0;
        q[ta++]=0;
        while(he!=ta)
        {
            i=q[he++];
            if(i+jump[i]>=n)break;
            if(!~d[i+jump[i]])d[q[ta++]=i+jump[i]]=d[i]+1;  / 等同于if(d[i+jump[i]]==-1)
            for(;l<i;l++)if(!~d[l])d[q[ta++]=l]=d[i]+1;
        }
        return d[i]+1;
    }
};