[leetcode]746. Min Cost Climbing Stairs

Solution 1:递归 tle

class Solution {
    public int minCostClimbingStairs(int[] cost) {
       
       int n=cost.length-1; 
       int start=-1;
        return help(cost,start,n);
    }
    
    public int help(int[] cost,int i,int n){
        if(i>n)return 0;
        if(i<0)return Math.min(help(cost,i+1,n),help(cost,i+2,n));;
      
        return cost[i]+Math.min(help(cost,i+1,n),help(cost,i+2,n));
    }
}

Solution 2: top down的dp

class Solution {
    public int minCostClimbingStairs(int[] cost) {
       
       int n=cost.length-1; 
       int start=-1;
       int[] dp=new int[n+1];
        Arrays.fill(dp,-1);
        return help(cost,start,n,dp);
    }
    
    public int help(int[] cost,int i,int n,int[] dp){
        
        if(i>n)return 0;
        if(i<0)return Math.min(help(cost,i+1,n,dp),help(cost,i+2,n,dp));;
      
        if(dp[i]>0)return dp[i];
        
        dp[i]=cost[i]+Math.min(help(cost,i+1,n,dp),help(cost,i+2,n,dp));
        return dp[i];
    }
}

Solution 3: bottom up的dp

今天突然豁然开朗了top down和bottom up
top down就是最直接的递归,一层一层下去。
bottom up就是反着来

class Solution {
    public int minCostClimbingStairs(int[] cost) {
       
       int n=cost.length-1; 
       int[] dp=new int[n+3];
        dp[n+1]=0;
        dp[n+2]=0;
        
        for(int i=n;i>=0;i--){
            dp[i]=cost[i]+Math.min(dp[i+1],dp[i+2]);
        }
        
        return Math.min(dp[0],dp[1]);
        
    }
    
  
}

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转载自blog.csdn.net/weixin_36869329/article/details/85003331