添加 problem 414

Algorithms/0414. Third Maximum Number/414. Third Maximum Number.go

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+package leetcode
+
+import (
+	"math"
+)
+
+func thirdMax(nums []int) int {
+	a, b, c := math.MinInt64, math.MinInt64, math.MinInt64
+	for _, v := range nums {
+		if v > a {
+			c = b
+			b = a
+			a = v
+		} else if v < a && v > b {
+			c = b
+			b = v
+		} else if v < b && v > c {
+			c = v
+		}
+	}
+	if c == math.MinInt64 {
+		return a
+	}
+	return c
+}

Algorithms/0414. Third Maximum Number/414. Third Maximum Number_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question414 struct {
+	para414
+	ans414
+}
+
+// para 是参数
+// one 代表第一个参数
+type para414 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans414 struct {
+	one int
+}
+
+func Test_Problem414(t *testing.T) {
+
+	qs := []question414{
+
+		question414{
+			para414{[]int{1, 1, 2}},
+			ans414{2},
+		},
+
+		question414{
+			para414{[]int{3, 2, 1}},
+			ans414{1},
+		},
+
+		question414{
+			para414{[]int{1, 2}},
+			ans414{2},
+		},
+
+		question414{
+			para414{[]int{2, 2, 3, 1}},
+			ans414{1},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 414------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans414, q.para414
+		fmt.Printf("【input】:%v       【output】:%v\n", p, thirdMax(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/0414. Third Maximum Number/README.md

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+# [414. Third Maximum Number](https://leetcode.com/problems/third-maximum-number/)
+
+## 题目
+
+Given a **non-empty** array of integers, return the **third** maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).
+
+**Example 1:**
+
+    Input: [3, 2, 1]
+    
+    Output: 1
+    
+    Explanation: The third maximum is 1.
+
+**Example 2:**
+
+    Input: [1, 2]
+    
+    Output: 2
+    
+    Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
+
+**Example 3:**
+
+    Input: [2, 2, 3, 1]
+    
+    Output: 1
+    
+    Explanation: Note that the third maximum here means the third maximum distinct number.
+    Both numbers with value 2 are both considered as second maximum.
+
+
+## 题目大意
+
+给定一个非空数组，返回此数组中第三大的数。如果不存在，则返回数组中最大的数。要求算法时间复杂度必须是 O(n)。
+
+
+## 解题思路
+
+- 水题，动态维护 3 个最大值即可。注意数组中有重复数据的情况。如果只有 2 个数或者 1 个数，则返回 2 个数中的最大值即可。