56.Merge Intervals

Tags: Medium Array Sort

Links: https://leetcode.com/problems/merge-intervals/

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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.

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class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int>> res;
        if (intervals.empty() || intervals[0].empty()) return res;

        sort(intervals.begin(), intervals.end());
        intervals.push_back({INT_MAX, INT_MAX});
        int tmpLeft = intervals[0][0], tmpRight = intervals[0][1];
        for (auto e : intervals) {
            if (e[0] > tmpRight) {
                res.push_back({tmpLeft, tmpRight});
                tmpLeft = e[0];
                tmpRight = e[1];
            }
            else {
                tmpRight = (e[1] > tmpRight) ? e[1] : tmpRight;
            }
        }

        return res;
    }
};

思路和SJTU OJ 1025基本一致，但是在数据结构上的选取略有区别，这是因为求解的目标不一致。因为SJTU OJ 1025求解的是区间长度，这里是需要保留最后的区间。

处理思路：

• 首先应该优先思考边界条件，即传入的intervals为空，优先处理。

• 考虑一般的情况，那么肯定要让左边界有序，触发推入结果数组res的条件是当前的左端点大于已知的右边界tmpRight。那么出问题也是在这里，因为可能存在两种情况：

在处理最后一个区间之前，前面已经得到了tmpLeft, tmpRight，最后一个区间为[a, b]：

• 如果a > tmpRight，触发推入结果数组条件，但是很显然最后一个区间无法假如结果数组
• 如果a <= tmpRight，不触发推入条件，只是更新了右边界，仍然会使结果少一个区间。

所以解决的方案是在intervals假如一个区间`[INT_MAX, INT_MAX]，保证触发推入条件，把最后一个区间推入结果数组。

方法二：

仍然是更新边界的思想，但是略微修改推入结果数组的触发条件，那么就无需要额外在intervals里加入一个区间了：

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        if (intervals.empty()) return {};
        sort(intervals.begin(), intervals.end());

        vector<vector<int>> res{intervals[0]};
        for (int i = 1; i < intervals.size(); ++i) {
            if (res.back()[1] < intervals[i][0]) {
                res.push_back(intervals[i]);
            } else {
                res.back()[1] = max(res.back()[1], intervals[i][1]);
            }
        }   
        return res;
    }
};