LEETCODE: Insert Interval

最新推荐文章于 2019-02-17 21:07:00 发布

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#LEETCODE #算法 #排序插入

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本文介绍了一种算法，用于在一个已排序的区间集合中插入一个新的区间，并在必要时进行合并操作。通过实例展示了如何处理区间重叠的情况。

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].

本来想用排序，再合并，但是这样就和上一次的方法一样了。用不同的方法，倒是费了九牛二虎之力。。。

/**
 * 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 {
    bool overlap(Interval ii, Interval jj) {
        if(ii.end < jj.start || ii.start > jj.end) {
            return false;
        }
        return true;
    }
    void merge(Interval ii, Interval& jj) {
        jj.start = ii.start < jj.start ? ii.start : jj.start;
        jj.end = ii.end > jj.end ? ii.end : jj.end;
    }
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval> results;
        if(intervals.size() == 0) {
            results.push_back(newInterval);
            return results;
        }
        bool inserted = false;
        for(int ii = 0; ii < intervals.size(); ii ++) {
            if(overlap(intervals[ii], newInterval)) {
                while(ii < intervals.size() && overlap(intervals[ii], newInterval)) {
                    merge(intervals[ii++], newInterval);
                }
                results.push_back(newInterval);
                ii --;
                inserted = true;
            }
            else {
                if(!inserted && newInterval.end < intervals[ii].start) {
                    results.push_back(newInterval);
                    inserted = true;
                }
                    
                results.push_back(intervals[ii]);
            }
        }
        if(!inserted && newInterval.start > intervals[intervals.size() - 1].end)
            results.push_back(newInterval);
        return results;
    }
};