leetcode: Insert Interval

最新推荐文章于 2021-02-09 11:10:51 发布

martin_liang

最新推荐文章于 2021-02-09 11:10:51 发布

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分类专栏： leetcode C++/C

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本文介绍了一个高效的算法，用于在一个已排序的非重叠区间集合中插入一个新的区间，并在必要时进行合并。通过两个实例展示了算法如何工作：第一个例子将[2,5]插入到[1,3],[6,9]中；第二个例子则演示了如何处理更复杂的场景，例如将[4,9]插入到[1,2],[3,5],[6,7],[8,10],[12,16]中。

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    int min(int a, int b)
    {
        if (a <= b)
            return a;
        return b;
    }
    
    int max(int a, int b)
    {
        if (a >= b)
            return a;
        return b;
    }

    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        
        vector<Interval> retVtr;
        
        int vtrSize = intervals.size();
        
        if (vtrSize == 0)
        {
            retVtr.push_back(newInterval);
            return retVtr;
        }
        
        int idx = 0;
        while (idx < vtrSize)
        {
            if (intervals[idx].end < newInterval.start)
            {
                retVtr.push_back(intervals[idx]);
                idx++;
            }
            else
            {
                break;
            }
        }
        
        int newStart = newInterval.start;
        int newEnd   = newInterval.end;
        
        while (idx < vtrSize)
        {
            //overlap
            if (intervals[idx].start <= newInterval.end)
            {
                newStart = min(newStart, intervals[idx].start);
                newEnd   = max(newEnd, intervals[idx].end);
                idx++;
            }
            else
            {
                break;
            }
        }
        
        retVtr.push_back(Interval(newStart, newEnd));
        
        while (idx < vtrSize)
        {
            retVtr.push_back(intervals[idx]);
            idx++;
        }
        
        return retVtr;
    }
};