leetcode 56. Merge Intervals-优快云博客

博客围绕合并重叠区间问题展开，给出示例输入输出，如输入[[1,3],[2,6]等，输出合并后的区间。还介绍了解题思路，先按起始点排序，重叠时更新右端点，不重叠则将区间加入输出，最后要处理最后一个区间。

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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 considered overlapping.

让把重叠的interval合并，输出所有的interval

思路：
这种interval重叠的问题，首先就想到了按start point排序
然后重叠的话更新右端点，否则把区间加入输出

注意最后一个区间要在循环结束的时候加入输出

//36ms
    public int[][] merge(int[][] intervals) {
        if (intervals == null || intervals.length == 0) {
            return new int[][]{};
        }
        
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
        
        ArrayList<int[]> arr = new ArrayList<>();
        int left = intervals[0][0];
        int right = intervals[0][1];
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] <= right) {
                right = Math.max(right, intervals[i][1]);
            } else {
                int[] tmp = new int[]{left, right};
                arr.add(tmp);
                left = intervals[i][0];
                right = intervals[i][1];
            }
        }
        
        arr.add(new int[]{left,right});
        
        int[][] result = new int[arr.size()][2];
        
        return arr.toArray(result);
    }

或者以下版本，原理是一样的

//36ms
    public int[][] merge(int[][] intervals) {
        if (intervals == null || intervals.length == 0) {
            return new int[][]{};
        }
        
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
        
        ArrayList<int[]> arr = new ArrayList<>();
        
        for (int[] interval: intervals) {
            if (arr.isEmpty() || interval[0] > arr.get(arr.size() - 1)[1]) {
                arr.add(interval);
            } else {
                arr.get(arr.size() - 1)[1] = Math.max(arr.get(arr.size() - 1)[1], interval[1]);
            }
        }
        
        int[][] result = new int[arr.size()][2];
        
        return arr.toArray(result);
    }

    public int[][] merge(int[][] intervals) {
        if(intervals == null || intervals.length == 0) {
            return new int[][]{};
        }
        
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
        List<int[]> tmp = new ArrayList<>();
        int left = intervals[0][0];
        int right = intervals[0][1];
        tmp.add(new int[]{left, right});
        
        for(int i = 1; i < intervals.length; i++) {
            if(intervals[i][0] > tmp.get(tmp.size() - 1)[1]) {
                tmp.add(new int[]{intervals[i][0], intervals[i][1]});
            } else {
                left = tmp.get(tmp.size() - 1)[0];
                right = Math.max(intervals[i][1], tmp.get(tmp.size() - 1)[1]);
                tmp.set(tmp.size()-1, new int[]{left, right});
            }
        }
        
        int[][] result = new int[tmp.size()][2];
        result = tmp.toArray(result);
        
        return result;
    }