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MergeIntervals.java
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84 lines (81 loc) · 2.42 KB
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package com.yangchd.leetcode.medium;
import java.util.ArrayList;
import java.util.List;
/**
* @author yangchd 2018/10/12.
*
* 56. Merge Intervals
* 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 considerred overlapping.
*
*/
public class MergeIntervals {
public class Interval {
int start;
int end;
Interval() {
start = 0;
end = 0;
}
Interval(int s, int e) {
start = s;
end = e;
}
}
/**
* 将需要合并的画在数组上，最后从数组中取出最后结果
*/
class Solution {
public List<Interval> merge(List<Interval> intervals) {
int max = 0;
for (Interval interval : intervals) {
max = Math.max(max, interval.end);
}
int[] flags = new int[max + 1];
for (int i = 0; i <= max; i++) {
flags[i] = 0;
}
for (Interval interval : intervals) {
if (interval.start == interval.end) {
if (flags[interval.start] == 0) {
flags[interval.start] = 2;
}
} else {
for (int m = interval.start; m < interval.end; m++) {
flags[m] = 1;
}
}
}
List<Interval> res = new ArrayList<Interval>();
for (int n = 0; n <= max; n++) {
if (flags[n] == 2) {
res.add(new Interval(n, n));
continue;
}
if (flags[n] >= 1) {
int index = 0;
while (n + index <= max && flags[n + index] >= 1) {
if (flags[n + index] == 2) {
break;
} else {
index++;
}
}
Interval interval = new Interval(n, n + index);
res.add(interval);
n = n + index;
}
}
return res;
}
}
}