[LeetCode] 56. Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Example 2:

Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        if (intervals.empty()) return {};
        // 先对区间进行排序，以 start 的值从小到大来排序
        sort(intervals.begin(), intervals.end());
        vector<vector<int>> res{intervals[0]};
        // 首先把第一个区间存入结果中，然后从第二个开始遍历区间集，如果结果中最后一个区间和遍历的当前区间无重叠，直接将当前区间存入结果中，
        // 如果有重叠，将结果中最后一个区间的 end 值更新为结果中最后一个区间的 end 和当前 end 值之中的较大值
        for (int i = 1; i < intervals.size(); ++i) {
            if (res.back()[1] >= intervals[i][0]) {
                res.back()[1] = max(res.back()[1], intervals[i][1]);
            } else {
                res.push_back(intervals[i]);
            }
        }
        return res;
    }
};

class Solution {
public:
    static bool compareAsc(const vector<int> &value1, const vector<int> &value2) {
        return value1[0] < value2[0];
    }
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        if (intervals.empty()) return {};
        sort(intervals.begin(), intervals.end(), compareAsc);
        vector<vector<int>> res{intervals[0]};
        for (int i = 1; i < intervals.size(); ++i) {
            if (res.back()[1] >= intervals[i][0]) {
                res.back()[1] = max(res.back()[1], intervals[i][1]);
            } else {
                res.push_back(intervals[i]);
            }
        }
        return res;
    }

};

上边两个都是等效的。

sort 函数的坑：