一刷436

Author: diguage

README.adoc

@@ -3078,12 +3078,12 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 |Medium
 |
 
-//|{counter:codes}
-//|{leetcode_base_url}/find-right-interval/[436. Find Right Interval^]
-//|{source_base_url}/_0436_FindRightInterval.java[Java]
-//|{doc_base_url}/0436-find-right-interval.adoc[题解]
-//|Medium
-//|
+|{counter:codes}
+|{leetcode_base_url}/find-right-interval/[436. Find Right Interval^]
+|{source_base_url}/_0436_FindRightInterval.java[Java]
+|{doc_base_url}/0436-find-right-interval.adoc[题解]
+|Medium
+|
 
 |{counter:codes}
 |{leetcode_base_url}/path-sum-iii/[437. Path Sum III^]

docs/0436-find-right-interval.adoc

@@ -1,65 +1,53 @@
 [#0436-find-right-interval]
-= 436. Find Right Interval
+= 436. 寻找右区间
 
-{leetcode}/problems/find-right-interval/[LeetCode - Find Right Interval^]
+https://leetcode.cn/problems/find-right-interval/[LeetCode - 436. 寻找右区间^]
 
-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.
+给你一个区间数组 `intervals`，其中 `intervals[i] = [start~i~, end~i~]`，且每个 `start~i~` 都 *不同* 。
 
-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.
+区间 `i` 的 *右侧区间* 可以记作区间 `j` ，并满足 `start~j~ >= end~i~` ，且 `start~j~` *最小化* 。注意 `i` 可能等于 `j` 。
 
-*Note:*
+返回一个由每个区间 `i` 的 *右侧区间* 在 `intervals` 中对应下标组成的数组。如果某个区间 `i` 不存在对应的 *右侧区间*，则下标 `i` 处的值设为 `-1` 。
 
+*示例 1：*
 
-. You may assume the interval's end point is always bigger than its start point.
-. You may assume none of these intervals have the same start point.
+....
+输入：intervals = [[1,2]]
+输出：[-1]
+解释：集合中只有一个区间，所以输出-1。
+....
 
+*示例 2：*
 
- 
+....
+输入：intervals = [[3,4],[2,3],[1,2]]
+输出：[-1,0,1]
+解释：对于 [3,4] ，没有满足条件的“右侧”区间。
+对于 [2,3] ，区间[3,4]具有最小的“右”起点;
+对于 [1,2] ，区间[2,3]具有最小的“右”起点。
+....
 
-*Example 1:*
+*示例 3：*
 
-[subs="verbatim,quotes,macros"]
-----
-*Input:* [ [1,2] ]
-
-*Output:* [-1]
-
-*Explanation:* There is only one interval in the collection, so it outputs -1.
-----
-
- 
+....
+输入：intervals = [[1,4],[2,3],[3,4]]
+输出：[-1,2,-1]
+解释：对于区间 [1,4] 和 [3,4] ，没有满足条件的“右侧”区间。
+对于 [2,3] ，区间 [3,4] 有最小的“右”起点。
+....
 
-*Example 2:*
-
-[subs="verbatim,quotes,macros"]
-----
-*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.
-----
+*提示：*
 
- 
+* `1 \<= intervals.length \<= 2 * 10^4^`
+* `+intervals[i].length == 2+`
+* `-10^6^ \<= start~i~ \<= end~i~ \<= 10^6^`
+* 每个间隔的起点都 *不相同*
 
-*Example 3:*
-
-[subs="verbatim,quotes,macros"]
-----
-*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.
-----
-
-*NOTE:* input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
 
 == 思路分析
 
+排序+二分查找。由于左端点都不同，可以根据左端点的值来来对区间进行排序。然后在这个有序序列中，查找右端点的“最近左端点”。
+
 [[src-0436]]
 [tabs]
 ====
@@ -77,8 +65,14 @@ include::{sourcedir}/_0436_FindRightInterval.java[tag=answer]
 // --
 // [{java_src_attr}]
 // ----
-// include::{sourcedir}/_0005_LongestPalindromicSubstring_2.java[tag=answer]
+// include::{sourcedir}/_0436_FindRightInterval_2.java[tag=answer]
 // ----
 // --
 ====
 
+
+== 参考资料
+
+. https://leetcode.cn/problems/find-right-interval/solutions/1506129/by-ac_oier-sijp/[436. 寻找右区间 - 一题双解 :「排序 + 二分」&「双指针（莫队思想）」^]
+. https://leetcode.cn/problems/find-right-interval/solutions/1504095/xun-zhao-you-qu-jian-by-leetcode-solutio-w2ic/[436. 寻找右区间 - 官方题解^]
+. https://leetcode.cn/problems/find-right-interval/solutions/52077/er-fen-cha-zhao-hong-hei-shu-by-liweiwei1419/[436. 寻找右区间 - 二分查找、二分搜索树^]

docs/index.adoc

@@ -957,7 +957,7 @@ include::0434-number-of-segments-in-a-string.adoc[leveloffset=+1]
 
 include::0435-non-overlapping-intervals.adoc[leveloffset=+1]
 
-// include::0436-find-right-interval.adoc[leveloffset=+1]
+include::0436-find-right-interval.adoc[leveloffset=+1]
 
 include::0437-path-sum-iii.adoc[leveloffset=+1]
 

logbook/202503.adoc

@@ -1349,6 +1349,10 @@ endif::[]
 |{doc_base_url}/0435-non-overlapping-intervals.adoc[题解]
 |❌ 贪心算法。换个角度，改为计算最多可以选多少个互不重叠的区间。
 
+|{counter:codes2503}
+|{leetcode_base_url}/find-right-interval/[436. 寻找右区间^]
+|{doc_base_url}/0436-find-right-interval.adoc[题解]
+|❌ 排序+二分查找。由于左端点都不同，可以根据左端点的值来来对区间进行排序。然后在这个有序序列中，查找右端点的“最近左端点”。
 
 |===
 

src/main/java/com/diguage/algo/leetcode/_0436_FindRightInterval.java

@@ -1,10 +1,41 @@
 package com.diguage.algo.leetcode;
 
+import java.util.Arrays;
+import java.util.Comparator;
+
 public class _0436_FindRightInterval {
   // tag::answer[]
   /**
    * @author D瓜哥 · https://www.diguage.com
-   * @since 2024-08-29 20:49:07
+   * @since 2025-08-16 16:16:17
    */
+  public int[] findRightInterval(int[][] intervals) {
+    int length = intervals.length;
+    Integer[] index = new Integer[length];
+    for (int i = 0; i < length; i++) {
+      index[i] = i;
+    }
+    Arrays.sort(index, Comparator.comparingInt((Integer a) -> intervals[a][0]));
+    int[] result = new int[length];
+    for (int i = 0; i < length; i++) {
+      int target = intervals[i][1];
+      int left = 0, right = length - 1;
+      while (left < right) {
+        int mid = left + (right - left) / 2;
+        int start = intervals[index[mid]][0];
+        if (start >= target) {
+          right = mid;
+        } else {
+          left = mid + 1;
+        }
+      }
+      result[i] = target <= intervals[index[left]][0] ? index[left] : -1;
+    }
+    return result;
+  }
   // end::answer[]
+  public static void main(String[] args) {
+    new _0436_FindRightInterval()
+      .findRightInterval(new int[][]{{3, 4}, {2, 3}, {1, 2}});
+  }
 }