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Find Right Interval (M)

题目

Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.

For any interval i, you need to store the minimum interval j‘s index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn‘t exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.

Note:

1. You may assume the interval‘s end point is always bigger than its start point.
2. You may assume none of these intervals have the same start point.

Example 1:

Input: [ [1,2] ]

Output: [-1]

Explanation: There is only one interval in the collection, so it outputs -1.

Example 2:

Input: [ [3,4], [2,3], [1,2] ]

Output: [-1, 0, 1]

Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.

Example 3:

Input: [ [1,4], [2,3], [3,4] ]

Output: [-1, 2, -1]

Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.

题意

给定一个区间集合S，对于其中的每一个区间s，判断能否在S中找到一个区间t，使得t的左端点大于等于s的右端点。如果能，则找到左端点最小的t。

思路

相当于给定两组数，一组为左端点的集合L，另一组为右端点的集合R，对于R中任意一个点a，在L中找到大于等于a的最小点b。使用二分法查找。

代码实现

Java

class Solution {
    public int[] findRightInterval(int[][] intervals) {
        int[] ans = new int[intervals.length];
      
      	// 将左端点抽出来和下标绑定，并排序
        List<int[]> indices = new ArrayList<>();
        for (int i = 0; i < intervals.length; i++) {
            indices.add(new int[] { intervals[i][0], i });
        }
        Collections.sort(indices, (int[] a, int[] b) -> a[0] - b[0]);
      
        for (int i = 0; i < intervals.length; i++) {
            int left = 0, right = indices.size() - 1;
            int target = intervals[i][1];
            while (left < right) {
                int mid = (right - left) / 2 + left;
                if (indices.get(mid)[0] < target) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            }
            if (indices.get(left)[0] >= target) {
                ans[i] = indices.get(left)[1];
            } else {
                ans[i] = -1;
            }
        }
      
        return ans;
    }
}

0436. Find Right Interval (M)

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原文地址：https://www.cnblogs.com/mapoos/p/13313669.html