Merge pull request #14 from rihib/search_insert_position

Search Insert Position

Author: rihib

pullrequests/search_insert_position/step1.go

@@ -0,0 +1,27 @@
+//lint:file-ignore U1000 Ignore all unused code
+package searchinsertposition
+
+/*
+時間：12分
+解法自体はすぐに浮かんで解くことができた。
+その割に二分探索を書くのがまだ慣れていなくて少し時間がかかってしまった。
+*/
+func searchInsert_step1(nums []int, target int) int {
+	left, right := 0, len(nums)
+	var mid int
+	for left < right {
+		mid = (left + right) / 2
+		if target == nums[mid] {
+			return mid
+		}
+		if target < nums[mid] {
+			right = mid
+		} else {
+			left = mid + 1
+		}
+	}
+	if target < nums[mid] {
+		return mid
+	}
+	return mid + 1
+}

pullrequests/search_insert_position/step2.go

@@ -0,0 +1,22 @@
+//lint:file-ignore U1000 Ignore all unused code
+package searchinsertposition
+
+/*
+最後の部分はleftを返せ馬良いと気づき変更した。
+その他もより読みやすいように直した。
+*/
+func searchInsert_step2(nums []int, target int) int {
+	left, right := 0, len(nums)
+	for left < right {
+		mid := (left + right) / 2
+		if nums[mid] == target {
+			return mid
+		}
+		if nums[mid] < target {
+			left = mid + 1
+		} else {
+			right = mid
+		}
+	}
+	return left
+}

pullrequests/search_insert_position/step3.go

@@ -0,0 +1,22 @@
+//lint:file-ignore U1000 Ignore all unused code
+package searchinsertposition
+
+/*
+オーバーフローを起こさないようにmidの計算方法を修正した
+https://github.com/hayashi-ay/leetcode/blob/064fca989bc4ecf9c7bce70237524a3e7ab1a21a/35.%20Search%20Insert%20Position.md?plain=1#L6
+*/
+func searchInsert_step3(nums []int, target int) int {
+	left, right := 0, len(nums)
+	for left < right {
+		mid := left + (right-left)/2
+		if nums[mid] == target {
+			return mid
+		}
+		if nums[mid] < target {
+			left = mid + 1
+		} else {
+			right = mid
+		}
+	}
+	return left
+}

pullrequests/search_insert_position/step4.go

@@ -0,0 +1,15 @@
+//lint:file-ignore U1000 Ignore all unused code
+package searchinsertposition
+
+func searchInsert(nums []int, target int) int {
+	left, right := 0, len(nums)
+	for left < right {
+		mid := left + (right-left)/2
+		if nums[mid] < target {
+			left = mid + 1
+		} else {
+			right = mid
+		}
+	}
+	return left
+}