• [LeetCode] Largest Number


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

    For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330.

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

    老题了,主要是搞懂cmp函数,前置0的处理还有待优化,不用copy两遍字符串。

     1 bool cmp(const string s1, const string s2) {
     2     return (s1 + s2) > (s2 + s1);
     3 }
     4 
     5 class Solution {
     6 public:
     7     string largestNumber(vector<int> &num) {
     8         vector<string> s_num(num.size());
     9         stringstream stream;
    10         for (int i = 0; i < num.size(); ++i) {
    11             stream << num[i];
    12             stream >> s_num[i];
    13             stream.clear();
    14         }
    15         sort(s_num.begin(), s_num.end(), cmp);
    16         string tmp_res;
    17         for (int i = 0; i < s_num.size(); ++i) {
    18             tmp_res += s_num[i];
    19         }
    20         string res;
    21         bool flag = false;
    22         for (int i = 0; i < tmp_res.size(); ++i) {
    23             if (tmp_res[i] != '0') {
    24                 res.push_back(tmp_res[i]);
    25                 flag = true;
    26             } else if (flag) {
    27                 res.push_back(tmp_res[i]);
    28             }
    29         }
    30         if (!flag) res.push_back('0');
    31         return res;
    32     }
    33 };

     因为字符串是从int转过来的,所以不会出现0123这样的数字,所以前置0只有一种情况,那就是“0”,所以处理前置0时只要考虑“0”这一种特殊情况就行了。

     1 class Solution {
     2 public:
     3     static bool cmp(const string &s1, const string &s2) {
     4         return (s1 + s2) > (s2 + s1);
     5     }
     6     string largestNumber(vector<int> &num) {
     7         vector<string> s_num(num.size());
     8         for (int i = 0; i < num.size(); ++i) {
     9             s_num[i] = to_string(num[i]);
    10         }
    11         sort(s_num.begin(), s_num.end(), cmp);
    12         string res;
    13         bool flag = false;
    14         for (int i = 0; i < s_num.size(); ++i) {
    15             if (s_num[i] != "0") {
    16                 res += s_num[i];
    17                 flag = true;
    18             } else if (flag) {
    19                 res += s_num[i];
    20             }
    21         }
    22         if (!flag) res.push_back('0');
    23         return res;
    24     }
    25 };
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  • 原文地址:https://www.cnblogs.com/easonliu/p/4221810.html
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