Jump gameII

leetcode jump gameII

看了题解，用BFS是比较好的思路，一层表示当前步能到的节点，curmax表示最远的，和贪心有异曲同工之妙

class Solution {
public:int jump(vector<int>& a) {int n=a.size();if(n<=1) return 0;int i=0, level=0, curmax=0, nextmax=0;while(i<=n-1){for(;i<=curmax && i<=n-1;i++){nextmax=max(nextmax, a[i]+i);if(nextmax>=n-1) return level+1;}curmax=nextmax;level++;}return -1;}
};

之前写的dp，TLE了，当DP思维训练了

class Solution {
public:int jump(vector<int>& a) {int n=a.size();if(n<=1) return 0;int dp[n];dp[0]=0;for(int i=1;i<n;i++){dp[i]=INT_MAX;for(int j=0;j<i;j++){if(a[j]>=i-j)dp[i]=min(dp[i], dp[j]+1);}}return dp[n-1];}
};

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