#184 Largest Number

题目描述：

Given a list of non negative integers, arrange them such that they form the largest number.

 Notice

The result may be very large, so you need to return a string instead of an integer.

Have you met this question in a real interview? 

Yes

Example

Given [1, 20, 23, 4, 8], the largest formed number is 8423201.

Challenge 

Do it in O(nlogn) time complexity.

题目思路：

这题的关键在于怎么定义compare函数，定义好了，直接把string按照排序好的vector写好就好了。

compare函数我是这么定义的（由大到小排序）：

1. 从两个string a和b的头开始找，如果发现a[idx_a] > b[idx_b]，就返回true

2. 如果a或者b的任何一个string已经找完，比如，b已经找完，但是a还没找完，不能简单的根据找完和没找完来判断（反例就是[2,20]和[2,23]）。应该把a没找完的char跟b的最后一个char比较：如果char a > char b end，那么还是a大([2,23])的例子，23应该比2大；反之就是[2,20]的例子，2比20大。对于a已经找完，但是b还没找完的情况也是类似的判断。

Mycode（AC = 15ms）：

class Solution {
public:
    /**
     *@param num: A list of non negative integers
     *@return: A string
     */
    string largestNumber(vector<int> &num) {
        // write your code here
        if (num.size() == 0) return "0";
        
        // define the compare function:
        // if a[idx_a] > b[idx_b], then a > b
        // after loop, 
        // if a is longer than b, then compare a[idx_a++] and b[idx_b - 1]
        // if b is longer than a, then compare a[idx_a - 1] and b[idx_b++]
        auto comp = [] (const string&a, const string&b) {
            int idx_a = 0, idx_b = 0;
            while (idx_a < a.length() && idx_b < b.length()) {
                if (a[idx_a] > b[idx_b]) {
                    return true;
                }
                else if (a[idx_a] < b[idx_b]) {
                    return false;
                }
                else {
                    idx_a++; idx_b++;
                }
            }
            
            //e.g., 2 compare with 23/20
            while (idx_a < a.length()) {
                if (a[idx_a] > b[idx_b - 1]) {
                    return true;
                }
                else if (a[idx_a] < b[idx_b - 1]){
                    return false;
                }
                idx_a++;
            }
            
            while (idx_b < b.length()) {
                if (b[idx_b] < a[idx_a - 1]) {
                    return true;
                }
                else if (b[idx_b] > a[idx_a - 1]) {
                    return false;
                }
                idx_b++;
            }
            
            return false;
        };
        
        // change the int to string, then do string sort
        vector<string> str_num(num.size(), "");
        for (int i = 0; i < str_num.size(); i++) {
            str_num[i] = to_string(num[i]);
        }
        
        sort(str_num.begin(), str_num.end(), comp);
        
        string ans = "";
        for (int i = 0; i < str_num.size(); i++) {
            ans += str_num[i];
        }
        
        // remove '0' at beginning
        while (ans.length() > 1 && ans[0] == '0') {
            ans = ans.substr(1);
        }
        
        return ans;
    }
};