Given a sorted integer array without duplicates, return the summary of its ranges.數組
For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].測試
題意:給出一個有序的整型數組且沒有重複項,返回它的範圍總結。
例如:數組 [0,1,2,4,5,7], 返回 ["0->2","4->5","7"]。code
個人作法是在原始數組後面添加一個哨兵‘0’,可是這個作法會致使末尾數是‘-1’時判斷失誤,所以須要作一下判斷,在數組最後一個數爲‘-1’時任意插入不爲‘0’的哨兵。string
class Solution { public: vector<string> summaryRanges(vector<int>& nums) { vector<string> range(0); if( nums.size() == 0 ){ return range; } vector<int>::iterator iter = nums.end() - 1; if( *iter == -1 ){ nums.push_back(2); }else{ nums.push_back(0); } iter = nums.begin(); for (vector<int>::iterator iter1 = nums.begin(); iter1 != nums.end() - 1; ++iter1){ if( *iter1 + 1 != *(iter1 + 1) ){ string rangeString = (iter == iter1)?(to_string( *iter)):(to_string(*iter)+"->"+to_string(*iter1)); range.push_back(rangeString); iter = iter1 + 1; } } return range; } };
如今代碼能夠經過測試了,但在我本身本地環境中沒法使用to_string()
方法,應該是類庫太老了,因此本身寫了一個。it
#include <sstream> string to_string(int n) { stringstream ss; ss << n; return ss.str(); }