leetCode(50):Summary Ranges 分类: leetCode 2015-07-24 11:17 122人阅读 评论(0) 收藏
Given a sorted integer array without duplicates, return the summary of its ranges.
For example, given [0,1,2,4,5,7]
, return ["0->2","4->5","7"].
class Solution { public: void intToString(int num, string& tmp) { bool flag = false; bool full = false; if (num < 0) { flag = true; } if (num == 0) tmp.push_back('0'); while (num!=0) { tmp.push_back('0' + abs(num % 10)); num = num / 10; } if (flag) {//最后压入负号 tmp.push_back('-'); }
//起初是将负数转换为正数,再取余,这样就不用取绝对值了(负数取余还是负数),但有一个问题,就是补码表示的正负数区间是不对称的,如-128是没问题的,可能128就不在表示范围之内了。<strong>所以不能直接将负数转换为正数</strong>。 reverse(tmp.begin() , tmp.end()); } vector<string> summaryRanges(vector<int>& nums) { vector<string> result; if (nums.empty()) return result; int tmpValue; string tmp; bool change = false; for (int i = 0; i < nums.size(); ++i) { if (tmp.empty()) { intToString(nums[i], tmp); tmpValue = nums[i]; change = false; } else { if (tmpValue + 1 == nums[i]) {//连续 tmpValue = nums[i]; change = true; } else {//不连续 if (change) {//有连续情况出现 tmp.push_back('-'); tmp.push_back('>'); string tmp1; intToString(tmpValue, tmp1); tmp+=tmp1; result.push_back(tmp); tmp.clear(); i--; } else {//没有连续的数 result.push_back(tmp); tmp.clear(); i--; } } } } if (change) {//最后一个string tmp.push_back('-'); tmp.push_back('>'); string tmp1; intToString(tmpValue, tmp1); tmp+=tmp1; result.push_back(tmp); tmp.clear(); } else { result.push_back(tmp); tmp.clear(); } return result; } };
网上看的别人的解法,很屌!
class Solution { public: vector<string> summaryRanges(vector<int>& nums) { vector<string> res; nums.push_back(INT_MIN); long long start = nums[0], last = nums[0]; for (int i = 1; i < nums.size(); i++) { if (last + 1 != (long long)nums[i]) { res.push_back(to_string(start)); if (last != start) {//to_string的使用,to_string可以将数值类型的变量(int,double,long等)转换为string类型 res[res.size() - 1] = res[res.size() - 1] + "->" + to_string(last); } start = nums[i]; } last = nums[i]; } return res; } };