添加 problem 441、457、781、984

Algorithms/0457. Circular Array Loop/457. Circular Array Loop.go

@@ -0,0 +1,37 @@
+package leetcode
+
+func circularArrayLoop(nums []int) bool {
+	if len(nums) == 0 {
+		return false
+	}
+	for i := 0; i < len(nums); i++ {
+		if nums[i] == 0 {
+			continue
+		}
+		// slow/fast pointer
+		slow, fast, val := i, getNextIndex(nums, i), 0
+		for nums[fast]*nums[i] > 0 && nums[getNextIndex(nums, fast)]*nums[i] > 0 {
+			if slow == fast {
+				// check for loop with only one element
+				if slow == getNextIndex(nums, slow) {
+					break
+				}
+				return true
+			}
+			slow = getNextIndex(nums, slow)
+			fast = getNextIndex(nums, getNextIndex(nums, fast))
+		}
+		// loop not found, set all element along the way to 0
+		slow, val = i, nums[i]
+		for nums[slow]*val > 0 {
+			next := getNextIndex(nums, slow)
+			nums[slow] = 0
+			slow = next
+		}
+	}
+	return false
+}
+
+func getNextIndex(nums []int, index int) int {
+	return ((nums[index]+index)%len(nums) + len(nums)) % len(nums)
+}

Algorithms/0457. Circular Array Loop/457. Circular Array Loop_test.go

@@ -0,0 +1,76 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question457 struct {
+	para457
+	ans457
+}
+
+// para 是参数
+// one 代表第一个参数
+type para457 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans457 struct {
+	one bool
+}
+
+func Test_Problem457(t *testing.T) {
+
+	qs := []question457{
+
+		question457{
+			para457{[]int{-1}},
+			ans457{false},
+		},
+
+		question457{
+			para457{[]int{3, 1, 2}},
+			ans457{true},
+		},
+
+		question457{
+			para457{[]int{-8, -1, 1, 7, 2}},
+			ans457{false},
+		},
+
+		question457{
+			para457{[]int{-1, -2, -3, -4, -5}},
+			ans457{false},
+		},
+
+		question457{
+			para457{[]int{}},
+			ans457{false},
+		},
+
+		question457{
+			para457{[]int{2, -1, 1, 2, 2}},
+			ans457{true},
+		},
+
+		question457{
+			para457{[]int{-1, 2}},
+			ans457{false},
+		},
+		question457{
+			para457{[]int{-2, 1, -1, -2, -2}},
+			ans457{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 457------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans457, q.para457
+		fmt.Printf("【input】:%v       【output】:%v\n", p, circularArrayLoop(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/0457. Circular Array Loop/README.md

@@ -0,0 +1,59 @@
+# [457. Circular Array Loop](https://leetcode.com/problems/circular-array-loop/)
+
+
+## 题目:
+
+You are given a **circular** array `nums` of positive and negative integers. If a number k at an index is positive, then move forward k steps. Conversely, if it's negative (-k), move backward k steps. Since the array is circular, you may assume that the last element's next element is the first element, and the first element's previous element is the last element.
+
+Determine if there is a loop (or a cycle) in `nums`. A cycle must start and end at the same index and the cycle's length > 1. Furthermore, movements in a cycle must all follow a single direction. In other words, a cycle must not consist of both forward and backward movements.
+
+**Example 1:**
+
+    Input: [2,-1,1,2,2]
+    Output: true
+    Explanation: There is a cycle, from index 0 -> 2 -> 3 -> 0. The cycle's length is 3.
+
+**Example 2:**
+
+    Input: [-1,2]
+    Output: false
+    Explanation: The movement from index 1 -> 1 -> 1 ... is not a cycle, because the cycle's length is 1. By definition the cycle's length must be greater than 1.
+
+**Example 3:**
+
+    Input: [-2,1,-1,-2,-2]
+    Output: false
+    Explanation: The movement from index 1 -> 2 -> 1 -> ... is not a cycle, because movement from index 1 -> 2 is a forward movement, but movement from index 2 -> 1 is a backward movement. All movements in a cycle must follow a single direction.
+
+**Note:**
+
+1. -1000 ≤ nums[i] ≤ 1000
+2. nums[i] ≠ 0
+3. 1 ≤ nums.length ≤ 5000
+
+**Follow up:**
+
+Could you solve it in **O(n)** time complexity and **O(1)** extra space complexity?
+
+
+## 题目大意
+
+给定一个含有正整数和负整数的环形数组 nums。 如果某个索引中的数 k 为正数，则向前移动 k 个索引。相反，如果是负数 (-k)，则向后移动 k 个索引。因为数组是环形的，所以可以假设最后一个元素的下一个元素是第一个元素，而第一个元素的前一个元素是最后一个元素。
+
+确定 nums 中是否存在循环（或周期）。循环必须在相同的索引处开始和结束并且循环长度 > 1。此外，一个循环中的所有运动都必须沿着同一方向进行。换句话说，一个循环中不能同时包括向前的运动和向后的运动。
+
+提示：
+
+- -1000 ≤ nums[i] ≤ 1000
+- nums[i] ≠ 0
+- 1 ≤ nums.length ≤ 5000
+
+进阶：
+
+- 你能写出时间时间复杂度为 O(n) 和额外空间复杂度为 O(1) 的算法吗？
+
+## 解题思路
+
+
+- 给出一个循环数组，数组的数字代表了前进和后退的步数，+ 代表往右(前进)，- 代表往左(后退)。问这个循环数组中是否存在一个循环，并且这个循环内不能只有一个元素，循环的方向都必须是同方向的。
+- 遇到循环就可以优先考虑用快慢指针的方法判断循环，这一题对循环增加了一个条件，循环不能只是单元素的循环，所以在快慢指针中加入这个判断条件。还有一个判断条件是循环的方向必须是同向的，这个简单，用 `num[i] * num[j] > 0` 就可以判断出是同向的(如果是反向的，那么两者的乘积必然是负数)，如果没有找到循环，可以将当前已经走过的路径上的 num[] 值都置为 0，标记已经访问过了。下次循环遇到访问过的元素，`num[i] * num[j] > 0` 就会是 0，提前退出找循环的过程。