添加 problem 969

Algorithms/969. Pancake Sorting/969. Pancake Sorting.go

@@ -0,0 +1,39 @@
+package leetcode
+
+func pancakeSort(A []int) []int {
+	if len(A) == 0 {
+		return []int{}
+	}
+	right := len(A)
+	var (
+		ans []int
+	)
+	for right > 0 {
+		idx := find(A, right)
+		if idx != right-1 {
+			reverse_(A, 0, idx)
+			reverse_(A, 0, right-1)
+			ans = append(ans, idx+1, right)
+		}
+		right--
+	}
+
+	return ans
+}
+
+func reverse_(nums []int, l, r int) {
+	for l < r {
+		nums[l], nums[r] = nums[r], nums[l]
+		l++
+		r--
+	}
+}
+
+func find(nums []int, t int) int {
+	for i, num := range nums {
+		if num == t {
+			return i
+		}
+	}
+	return -1
+}

Algorithms/969. Pancake Sorting/969. Pancake Sorting_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question969 struct {
+	para969
+	ans969
+}
+
+// para 是参数
+// one 代表第一个参数
+type para969 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans969 struct {
+	one []int
+}
+
+func Test_Problem969(t *testing.T) {
+
+	qs := []question969{
+
+		question969{
+			para969{[]int{}},
+			ans969{[]int{}},
+		},
+
+		question969{
+			para969{[]int{1}},
+			ans969{[]int{1}},
+		},
+
+		question969{
+			para969{[]int{3, 2, 4, 1}},
+			ans969{[]int{3, 4, 2, 3, 1, 2}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 969------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans969, q.para969
+		fmt.Printf("【input】:%v       【output】:%v\n", p, pancakeSort(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/969. Pancake Sorting/README.md

@@ -0,0 +1,43 @@
+# [969. Pancake Sorting](https://leetcode.com/problems/pancake-sorting/)
+
+## 题目
+
+Given an array A, we can perform a pancake flip: We choose some positive integer k <= A.length, then reverse the order of the first k elements of A.  We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A.
+
+Return the k-values corresponding to a sequence of pancake flips that sort A.  Any valid answer that sorts the array within 10 * A.length flips will be judged as correct.
+
+Example 1:
+
+```c
+Input: [3,2,4,1]
+Output: [4,2,4,3]
+Explanation: 
+We perform 4 pancake flips, with k values 4, 2, 4, and 3.
+Starting state: A = [3, 2, 4, 1]
+After 1st flip (k=4): A = [1, 4, 2, 3]
+After 2nd flip (k=2): A = [4, 1, 2, 3]
+After 3rd flip (k=4): A = [3, 2, 1, 4]
+After 4th flip (k=3): A = [1, 2, 3, 4], which is sorted. 
+```
+
+Example 2:
+
+```c
+Input: [1,2,3]
+Output: []
+Explanation: The input is already sorted, so there is no need to flip anything.
+Note that other answers, such as [3, 3], would also be accepted.
+```
+
+Note:
+
+- 1 <= A.length <= 100
+- A[i] is a permutation of [1, 2, ..., A.length]
+
+## 题目大意
+
+给定一个数组，要求输出“煎饼排序”的步骤，使得最终数组是从小到大有序的。“煎饼排序”，每次排序都反转前 n 个数，n 小于数组的长度。
+
+这道题的思路是，每次找到当前数组中无序段中最大的值，（初始的时候，整个数组相当于都是无序段），将最大值的下标 i 进行“煎饼排序”，前 i 个元素都反转一遍。这样最大值就到了第一个位置了。然后紧接着再进行一次数组总长度 n 的“煎饼排序”，目的是使最大值到数组最后一位，这样它的位置就归位了。那么数组的无序段为 n-1 。然后用这个方法不断的循环，直至数组中每个元素都到了排序后最终的位置下标上了。最终数组就有序了。
+
+这道题有一个特殊点在于，数组里面的元素都是自然整数，那么最终数组排序完成以后，数组的长度就是最大值。所以找最大值也不需要遍历一次数组了，直接取出长度就是最大值。