标签：dig 数据规模 mos content class nat input reverse pre
Given a non-negative integer, you could swap two digits at most once to get the maximum valued number. Return the maximum valued number you could get.
Example 1:
Input: 2736 Output: 7236 Explanation: Swap the number 2 and the number 7.
Example 2:
Input: 9973 Output: 9973 Explanation: No swap.
Note:
思路：
暴力枚举所有交换可能，竟然没有超时。。。感觉因为输入最多到108
最多8位，复杂度O(n3)虽然很高，但数据规模小，但总之不是什么好办法。
vector<int> digits(int num) { if (num == 0)return{0}; vector<int> ret; while (num > 0) { ret.push_back(num % 10); num /= 10; } reverse(ret.begin(), ret.end()); return ret; } int toNum(vector<int>num) { int r = 0; for (int i = 0; i < num.size();i++) { r *= 10; r += num[i]; } return r; } int maximumSwap(int n) { vector<int> num = digits(n); int m = n; for (int i = 0; i < num.size();i++) { for (int j = i + 1; j < num.size();j++) { swap(num[i], num[j]); m = max(m, toNum(num)); swap(num[i], num[j]); } } return m; }
标签：dig 数据规模 mos content class nat input reverse pre
原文地址：http://www.cnblogs.com/hellowooorld/p/7469326.html
