salamanderMH
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一道算法题,请教一下

  •  
  •   salamanderMH · Sep 6, 2020 · 1483 views
    This topic created in 2075 days ago, the information mentioned may be changed or developed.

    问题

    有砝码 1g,2g,3g...100g,组成 100g 的重量有几种方式?
    这道题应该可以用动态规划做,但一下子没想出来(太渣了) 写了一个回溯的算法,但效率太差了:

    function counterweightWays(currentNum, allNum, leftWeight, tmpResult, result) {    
        if (currentNum > allNum) {        
            return    }    
        if (leftWeight == 0) {        
            result.push(Array.from(tmpResult))        
            return    
        }    
        const maxNum = Math.floor(leftWeight / currentNum)    
        for (let n = maxNum; n >= 0; n--) {        
            tmpResult.push(n)        
            counterweightWays(currentNum + 1, allNum, leftWeight - n * currentNum, tmpResult, result)        
            tmpResult.pop()    
        }
    }
    
    5 replies    2020-09-07 09:09:59 +08:00
    jmc891205
        1
    jmc891205  
       Sep 6, 2020 via iPhone
    搜索一下零钱兑换问题
    fishCatcher
        2
    fishCatcher  
       Sep 6, 2020   ❤️ 1
    // 设 dp[i]是组成 i 克的方法个数, 1 <= i <= n
    for (int i = 1; i <= n; i++)
    dp[i] = 1; // base case,直接拿 1 个 i 克的砝码即可
    for (int i = 2; i <= n; i++)
    for (int j = 1; j <= i / 2; j++)
    dp[i] += dp[j] * dp[i - j];
    return dp[n];
    salamanderMH
        4
    salamanderMH  
    OP
       Sep 6, 2020
    salamanderMH
        5
    salamanderMH  
    OP
       Sep 7, 2020
    @zhy0216 哇,太强了。
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