剑指 Offer 14- I. 剪绳子

class Solution {
    public int cuttingRope(int n) {
        int[] dp = new int[n+1];
        dp[1] = 1;
        dp[2] = 1;
        for(int i = 3;i<=n;i++){
            int x1 = i/2;
            int x2 = i - x1;
            int result = Math.max(x1,dp[x1])*Math.max(x2,dp[x2]);
            while(x1>2){
                x1--;
                x2++;
                result = Math.max(result,Math.max(x1,dp[x1])*Math.max(x2,dp[x2]));
            }
            dp[i] = result;
        }
        return dp[n];
    }
}

 动态规划

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贪婪算法

每次剪出长度为3的绳子，当长度为4时，剪成长度为2和2的绳子。

public int cuttingRope(int n) {
        if(n<2) return 0;
        if(n == 2) return 1;
        if(n == 3) return 2;
        int times_of_3 = n/3;
        if(n%3 == 1){
            times_of_3 --;
        }
        int times_of_2 = (n-times_of_3*3)/2;
        return (int)(Math.pow(3,times_of_3)*Math.pow(2,times_of_2));
    }