题目

给定一个无重复元素的有序整数数组，返回数组区间范围的汇总。

示例 1:

输入: [0,1,2,4,5,7]
输出: ["0->2","4->5","7"]
解释: 0,1,2 可组成一个连续的区间; 4,5 可组成一个连续的区间。

示例 2:

输入: [0,2,3,4,6,8,9]
输出: ["0","2->4","6","8->9"]
解释: 2,3,4 可组成一个连续的区间; 8,9 可组成一个连续的区间。

C++解法

#include <iostream>
#include <vector>
#include <map>
#include <set>
using namespace std;
class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        vector<string> ranges;
        if (nums.empty()) return ranges;
        if (nums.size() == 1) {
            ranges.push_back(to_string(nums[0]));
            return ranges;
        }
        long lower = -1;
        long delta = 0;
        long upper = -1;
        for (int i = 0; i < nums.size() - 1; i++) {
            if (lower == -1) {
                lower = nums[i];
                upper = nums[i];
            }
            delta = (long)nums[i+1] - (long)nums[i];
            if (delta == 1) { upper = nums[i + 1]; }
            if (delta != 1 || i == nums.size() - 2) {
                if (lower == upper) {
                    ranges.push_back(to_string(lower));
                } else {
                    ranges.push_back(to_string(lower) + "->" + to_string(upper));
                }
                if (upper != nums[i + 1] && i == nums.size() - 2) ranges.push_back(to_string(nums[i + 1]));
                lower = -1;
                upper = -1;
            }
        }
        return ranges;
    }
};

int main(int argc, const char * argv[]) {
    // insert code here...
    Solution solution;
    vector<int> vec {0,1,2,4,5,7};
    auto ranges = solution.summaryRanges(vec);
    for (auto range: ranges) {
        cout << range << endl;
    }
    vec = {0, 2, 3, 4, 6, 8, 9};
    ranges = solution.summaryRanges(vec);
    for (auto range: ranges) {
        cout << range << endl;
    }
    vec = {-2147483648,-2147483647,2147483647};
    ranges = solution.summaryRanges(vec);
    for (auto range: ranges) {
        cout << range << endl;
    }
    return 0;
}

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/summary-ranges