697 accepted. 108ms > 55ms

Author: aQua

Algorithms/0697.degree-of-an-array/README.md

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+# [697. Degree of an Array](https://leetcode.com/problems/degree-of-an-array/)
+
+## 题目
+
+Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.
+
+Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.
+
+Example 1:
+
+```text
+Input: [1, 2, 2, 3, 1]
+Output: 2
+Explanation:
+The input array has a degree of 2 because both elements 1 and 2 appear twice.
+Of the subarrays that have the same degree:
+[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
+The shortest length is 2. So return 2.
+```
+
+Example 2:
+
+```text
+Input: [1,2,2,3,1,4,2]
+Output: 6
+```
+
+Note:
+
+- nums.length will be between 1 and 50,000.
+- nums[i] will be an integer between 0 and 49,999.
+
+## 解题思路
+
+见程序注释

Algorithms/0697.degree-of-an-array/degree-of-an-array.go

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+package Problem0697
+
+func findShortestSubArray(nums []int) int {
+	size := len(nums)
+	if size < 2 {
+		return size
+	}
+
+	first := make(map[int]int, size)
+	count := make(map[int]int, size)
+	maxCount := 1
+	minLen := size
+	for i, n := range nums {
+		count[n]++
+		if count[n] == 1 {
+			first[n] = i
+		} else {
+			l := i - first[n] + 1
+			if maxCount < count[n] ||
+				(maxCount == count[n] && minLen > l) {
+				maxCount = count[n]
+				minLen = l
+			}
+		}
+	}
+
+	if len(count) == size {
+		return 1
+	}
+	return minLen
+}

Algorithms/0697.degree-of-an-array/degree-of-an-array_test.go

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+package Problem0697
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/stretchr/testify/assert"
+)
+
+// tcs is testcase slice
+var tcs = []struct {
+	nums []int
+	ans  int
+}{
+
+	{
+		[]int{1, 2, 2, 3, 1},
+		2,
+	},
+
+	{
+		[]int{1, 2, 2, 3, 1, 4, 2},
+		6,
+	},
+
+	// 可以有多个 testcase
+}
+
+func Test_fn(t *testing.T) {
+	ast := assert.New(t)
+
+	for _, tc := range tcs {
+		fmt.Printf("~~%v~~\n", tc)
+		ast.Equal(tc.ans, findShortestSubArray(tc.nums), "输入:%v", tc)
+	}
+}
+
+func Benchmark_fn(b *testing.B) {
+	for i := 0; i < b.N; i++ {
+		for _, tc := range tcs {
+			findShortestSubArray(tc.nums)
+		}
+	}
+}