From dd6fab41a3cf9c95d06ffa7a05a6a9c0744d43a6 Mon Sep 17 00:00:00 2001
From: yamashita-ki <yamashita.kiyoshi.rac@gmail.com>
Date: Wed, 1 Oct 2025 00:34:40 +0900
Subject: [PATCH] 704. Binary Search

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 704. Binary Search.md | 135 ++++++++++++++++++++++++++++++++++++++++++
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 create mode 100644 704. Binary Search.md

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+## step1 とりあえず解く
+- 計算量
+  - 時間:O(N)
+  - 空間:O(1)
+- まず思いついた全探索
+
+```java
+class Solution {
+    public int search(int[] nums, int target) {
+        for (int i = 0; i < nums.length; i++) {
+            if (nums[i] == target) {
+                return i;
+            }
+        }
+        return -1;
+    }
+}
+
+```
+
+## step2 他の人の回答を見る
+- 計算量
+  - 時間:O(logN)
+  - 空間:O(1)
+- ビルドインで提供されているメソッドを使う
+    - 「キーが見つかった場合にのみ戻り値が>= 0になることが保証される。」らしい
+    - https://docs.oracle.com/javase/jp/8/docs/api/java/util/Arrays.html#binarySearch-int:A-int-
+- 面接でライブラリの使用が許されていないかも？
+
+```java
+public class Solution {
+    public int search(int[] nums, int target) {
+        int index = Arrays.binarySearch(nums, target);
+        return index >= 0 ? index : -1;
+    }
+}
+
+```
+
+- 計算量
+  - 時間:O(logN)
+  - 空間:O(1)
+- 一番よく見る二分探索
+
+```java
+public class Solution {
+    public int search(int[] nums, int target) {
+        int left = 0, right = nums.length - 1;
+
+        while (left <= right) {
+            int middle = left + ((right - left) / 2);
+            if (nums[middle] > target) {
+                right = middle - 1;
+            } else if (nums[middle] < target) {
+                left = middle + 1;
+            } else {
+                return middle;
+            }
+        }
+        return -1;
+    }
+}
+
+```
+
+- 計算量
+  - 時間:O(logN)
+  - 空間:O(1)
+- Upper Bound
+- targetを超える最初のindexを返す、そのためreturnで調整している
+
+```java
+public class Solution {
+    public int search(int[] nums, int target) {
+        int left = 0, right = nums.length;
+
+        while (left < right) {
+            int middle = left + ((right - left) / 2);
+            if (nums[middle] > target) {
+                right = middle;
+            } else {
+                left = middle + 1;
+            }
+        }
+        return (left > 0 && nums[left - 1] == target) ? left - 1 : -1;
+    }
+}
+```
+
+- 計算量
+  - 時間:O(logN)
+  - 空間:O(1)
+- Lower Bound
+- target以上の最初の要素を探す
+
+```java
+public class Solution {
+    public int search(int[] nums, int target) {
+        int left = 0, right = nums.length;
+
+        while (left < right) {
+            int middle = left + (right - left) / 2;
+            if (nums[middle] >= target) {
+                right = middle;
+            } else {
+                left = middle + 1;
+            }
+        }
+        return (left < nums.length && nums[left] == target) ? left : -1;
+    }
+}
+```
+
+## step3 3回とく
+- 今回の問題には古典的な二分探索が最適だと感じたのでそのまま
+
+```java
+class Solution {
+    public int search(int[] nums, int target) {
+        int left = 0, right = nums.length - 1;
+        while (left <= right) {
+            int middle = (left + right) / 2;
+            if (nums[middle] == target) {
+                return middle;
+            } else if (target > nums[middle]) {
+                left = middle + 1;
+            } else {
+                right = middle - 1;
+            }
+        }
+        return -1;
+    }
+}
+
+```
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