Add roman-to-integer

Author: beyondyyh

algorithms/README.md

@@ -29,6 +29,7 @@
 |0102|[二叉树的层次遍历](https://leetcode-cn.com/problems/binary-tree-level-order-traversal/submissions/)|[go](./tree/102.levelOrder.go)|M|
 |0103|[二叉树的锯齿形层次遍历](https://leetcode-cn.com/problems/binary-tree-level-order-traversal/submissions/)|[go](./tree/103.zigzagLevelOrder.go)|M|
 |0104|[二叉树的最大深度](https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/)|[go](./tree/104.maxDepth.go)|S|
+|0113|[罗马数字转整数](https://leetcode-cn.com/problems/roman-to-integer/)|[go](./mystring/113.romanToInt.go)|S|
 |0118|[杨辉三角](https://leetcode-cn.com/problems/pascals-triangle/)|[go](./myarray/118.generate.go)|S|
 |0121|[买卖股票的最佳时机](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/)|[go](./myarray/121.maxProfit.go)|S|
 |0136|[只出现一次的数字](https://leetcode-cn.com/problems/single-number/comments/)|[go](./mybit/136.singleNumber.go)|S|

algorithms/mystring/13.romanToInt.go

@@ -0,0 +1,75 @@
+package mystring
+
+// 思路：建立一个Map来映射符号和值，然后对字符串从后向前遍历，保留上一个字符串
+// 如果当前字符代表的值大于等于上一个（右边），就加上该值；
+// 如果当前字符代表的值小于上一个（右边），则减去该值。
+// romanToInt1 returns int
+func romanToInt1(s string) int {
+	m := map[byte]int{
+		'I': 1,
+		'V': 5,
+		'X': 10,
+		'L': 50,
+		'C': 100,
+		'D': 500,
+		'M': 1000,
+	}
+
+	var res, last int
+	for i := len(s) - 1; i >= 0; i-- {
+		curr := m[s[i]]
+		flag := 1
+		// 小数在大数的左边，要减去小数
+		if curr < last {
+			flag = -1
+		}
+
+		res += flag * curr
+		last = curr
+	}
+
+	return res
+}
+
+// 思路：将所有的组合可能性列出并添加到哈希表中，
+// 然后对字符串进行遍历，先判断两个字符的组合在哈希表中是否存在，存在则将值取出加到结果 res 中，并向后移2个字符；
+// 不存在则将判断当前 1 个字符是否存在，存在则将值取出加到结果 res 中，并向后移 1 个字符。
+// romanToInt2 returns int
+func romanToInt2(s string) int {
+	m := map[string]int{
+		"I":  1,
+		"V":  5,
+		"X":  10,
+		"L":  50,
+		"C":  100,
+		"D":  500,
+		"M":  1000,
+		"IV": 4,
+		"IX": 9,
+		"XL": 40,
+		"XC": 90,
+		"CD": 400,
+		"CM": 900,
+	}
+
+	var res int
+	for i := 0; i < len(s); {
+		if i+1 < len(s) && containsKey(m, string(s[i:i+2])) {
+			res += m[string(s[i:i+2])]
+			i += 2
+		} else if containsKey(m, string(s[i:i+1])) {
+			res += m[string(s[i:i+1])]
+			i++
+		}
+	}
+
+	return res
+}
+
+// containsKey return true if m contains k
+func containsKey(m map[string]int, k string) bool {
+	if _, ok := m[k]; ok {
+		return true
+	}
+	return false
+}

algorithms/mystring/13.romanToInt_test.go

@@ -0,0 +1,49 @@
+package mystring
+
+import (
+	"Leetcode/algorithms/kit"
+	"testing"
+)
+
+// run: go test -v 13.*
+func Test_romanToInt(t *testing.T) {
+	cases := []kit.CaseEntry{
+		{
+			Name:     "x1",
+			Input:    "III",
+			Expected: 3,
+		},
+		{
+			Name:     "x2",
+			Input:    "IV",
+			Expected: 4,
+		},
+		{
+			Name:     "x3",
+			Input:    "IX",
+			Expected: 9,
+		},
+		{
+			Name:     "x4",
+			Input:    "LVIII",
+			Expected: 58,
+		},
+		{
+			Name:     "x5",
+			Input:    "MCMXCIV",
+			Expected: 1994,
+		},
+	}
+
+	for _, c := range cases {
+		t.Run(c.Name, func(t *testing.T) {
+			if output := romanToInt1(c.Input.(string)); output != c.Expected.(int) {
+				t.Errorf("romanToInt1(%s)=%d, expected=%d", c.Input, output, c.Expected)
+			}
+
+			if output := romanToInt2(c.Input.(string)); output != c.Expected.(int) {
+				t.Errorf("romanToInt2(%s)=%d, expected=%d", c.Input, output, c.Expected)
+			}
+		})
+	}
+}