LeetCode 56. Merge Intervals

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本文介绍了一种有效的合并区间算法，该算法首先将输入的区间按起始值进行排序，然后通过迭代的方式合并所有重叠的区间。具体实现包括两种方法：一种使用额外的空间存储结果，另一种则采用原地修改的方式，利用双指针技术完成合并。

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

The basic idea is to sort the intervals increasingly based on the .start value. Then, merge them one by one until merge fails.

static bool sorter(Interval start, Interval end) {
    return start.start < end.start;
}

static bool merge(Interval& v1, Interval& v2, Interval& res) {
    if(v1.end < v2.start) return false;
    res.start = v1.start;
    res.end = max(v1.end, v2.end);
    return true;
}

vector<Interval> merge(vector<Interval>& intervals) {
    if(intervals.size() == 0) return {};
    sort(intervals.begin(), intervals.end(), sorter);
    vector<Interval> res;
    res.push_back(intervals[0]);
    Interval tmp;
    for(int i = 1; i < intervals.size(); ++i) {
        if(merge(res.back(), intervals[i], tmp)) {
            res.back() = tmp;
        } else {
            res.push_back(intervals[i]);
        }
    }
    return res;
}

Do it in-place, using two pointers.

static bool sorter(Interval a, Interval b) {
  return a.start <= b.start;
}

void mergeInterval(vector<Interval>& nums) {
  if(nums.size() <= 1) return;
  int index = 0;
  sort(nums.begin(), nums.end(), sorter);
  for(int i = 0; i < nums.size(); ++i) {
    if(index != 0 && nums[index - 1].end >= nums[i].start) {
      while(index != 0 && nums[index-1].end >= nums[i].start) {
        nums[index-1].start = min(nums[index-1].start, nums[i].start);
        nums[index-1].end = max(nums[index-1].end, nums[i].end);
        index--;
      }
    } else {
      nums[index] = nums[i];
    }
    index++;
  }
}