【Leetcode 553 】Optimal Division

问题描述：

Given a list of positive integers, the adjacent integers will perform the float division. For example, [2,3,4] -> 2 / 3 / 4.

However, you can add any number of parenthesis at any position to change the priority of operations. You should find out how to add parenthesis to get the maximum result, and return the corresponding expression in string format. Your expression should NOT contain redundant parenthesis.

 

解法：

一串数字之间都是除号，要加括号使得它们的结果最大。

这道题不是很好的算法题，用tricky的方法一下子就解出来了，只要加一个括号就行了。但是仍然可以用分治法去求解这个问题，只是代码的量比较大，结果也一样。

Solution如下：

class Solution {
public:
    string optimalDivision(vector<int>& nums) {

        if (nums.size() == 1) {
            return to_string(nums[0]);
        }
        if (nums.size() == 2) {
            return to_string(nums[0]) + "/" + to_string(nums[1]);
        }
        string s = to_string(nums[0]) + "/";
        string t = to_string(nums[1]);
        for (int i = 2; i < nums.size(); i++) {
            t += "/" + to_string(nums[i]);
        }
        s = s + "(" + t + ")";
        return s;
    }
    string to_string(int n) {
        const int maxsize = 20;
        char c[maxsize];
        snprintf(c, maxsize, "%d\0", n);
        string s(c);
        return s;
    }
};

 

分治法Solution：

public class Solution {
    class Result {
        String str;
        double val;
    }
    
    public String optimalDivision(int[] nums) {
        int len = nums.length;
        return getMax(nums, 0, len - 1).str;
    }
    
    private Result getMax(int[] nums, int start, int end) {
        Result r = new Result();
        r.val = -1.0;
        
        if (start == end) {
            r.str = nums[start] + "";
            r.val = (double)nums[start];
        }
        else if (start + 1 == end) {
            r.str = nums[start] + "/" + nums[end];
            r.val = (double)nums[start] / (double)nums[end];
        }
        else {
            for (int i = start; i < end; i++) {
                Result r1 = getMax(nums, start, i);
                Result r2 = getMin(nums, i + 1, end);
                if (r1.val / r2.val > r.val) {
                    r.str = r1.str + "/" + (end - i >= 2 ? "(" + r2.str + ")" : r2.str);
                    r.val = r1.val / r2.val;
                }
            }
        }
        
        //System.out.println("getMax " + start + " " + end + "->" + r.str + ":" + r.val);
        return r;
    }
    
    private Result getMin(int[] nums, int start, int end) {
        Result r = new Result();
        r.val = Double.MAX_VALUE;
        
        if (start == end) {
            r.str = nums[start] + "";
            r.val = (double)nums[start];
        }
        else if (start + 1 == end) {
            r.str = nums[start] + "/" + nums[end];
            r.val = (double)nums[start] / (double)nums[end];
        }
        else {
            for (int i = start; i < end; i++) {
                Result r1 = getMin(nums, start, i);
                Result r2 = getMax(nums, i + 1, end);
                if (r1.val / r2.val < r.val) {
                    r.str = r1.str + "/" + (end - i >= 2 ? "(" + r2.str + ")" : r2.str);
                    r.val = r1.val / r2.val;
                }
            }
        }
        
        //System.out.println("getMin " + start + " " + end + "->" + r.str + ":" + r.val);
        return r;
    }
}