LeetCode Insert Interval

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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].

思路分析：这题题目本身并不难，但是还是调试了很久。定义结果res容器，首先将newInterval的start和intervals里面区间的end比较，找到应该插入newInterval的位置，插入newInterval有点讲究，应该先添加newInterval（添加前可能需要更新其start）到res，然后继续将后检查，看添加的newInterval是否还与后面的区间有冲突，如果有冲突，再更新newInterval的end一个一个合并后面的可以merge的区间（条件是newInterval.end >= intervals.get(i).start）。在java里面，List容器里面保存的是对象newInterval的引用（地址），所以，对newInterval的修改也相当于对res的修改。这样向后检查一直到merge完所有可以merge的区间，最后添加剩下的区间到res即可。时间复杂度和空间复杂度都是O（N）。实现的时候，可以用一个指针i从头到尾遍历intervals到底，针对不同的段进行不同处理。这题和LeetCode Merge Intervals有关系，可以联系起来思考。

AC Code

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        //1120
        List<Interval> res = new ArrayList<Interval>();
        if(intervals.size() == 0) {
            res.add(newInterval);
            return res;
        }
        int i = 0;
        while(i < intervals.size() && newInterval.start > intervals.get(i).end){
            res.add(intervals.get(i));
            i++;
        }
        if(i < intervals.size())
            newInterval.start = Math.min(newInterval.start, intervals.get(i).start);
        res.add(newInterval);
        //first add newInterval, then use newInterval to merge the afterwards intervals that within the range
        //Since we mantain reference of newInterval in res, the modification of newInterval also modify res
        //merge afterwards intervals one by one
        while(i < intervals.size() && newInterval.end >= intervals.get(i).start){
            newInterval.end = Math.max(newInterval.end, intervals.get(i).end);
            i++;
        }
        for(int j = i; j < intervals.size(); j++){
            res.add(intervals.get(j));
        }
        return res;
        //1139
    }
}

LeetCode Insert Interval

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原文地址：http://blog.csdn.net/yangliuy/article/details/44408743