题目:
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”].
题意:
给定一个有序的无重复元素的数组,合并它们的区间,比如0,1,2就是区间0->2。
思路:
采用两个指针,一个指向区间的头元素,一个从该元素之后往后扫描,如果扫描的元素都比前一个大1,那继续扫描。
以上。
代码如下:
class Solution {
public:
vector<string> summaryRanges(vector<int>& nums) {
vector<string> result;
int size = nums.size();
if (size == 0)return result;
else if(size == 1) {
stringstream ss;
ss << nums[0];
result.push_back(ss.str());
return result;
}
int left = 0;
int right = 1;
while (right < size) {
while (right < size && (nums[right] == nums[right - 1] + 1))right++;
stringstream ss;
ss << nums[left];
if (right - 1 != left) {
ss << "->" << nums[right - 1];
}
result.push_back(ss.str());
left = right;
right++;
if (right == size) {
stringstream ss;
ss << nums[left];
if (right - 1 != left) {
ss << "->" << nums[right - 1];
}
result.push_back(ss.str());
break;
}
}
return result;
}
};