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Commit 0eaafc5

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Merge pull request doocs#228 from ElectricBubble/master
Add solution 13, 21, 27, 28, 35, 38, 53, 58 in Golang
2 parents 12aac43 + 2c85704 commit 0eaafc5

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10 files changed

+185
-0
lines changed

10 files changed

+185
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Lines changed: 14 additions & 0 deletions
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func romanToInt(s string) int {
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symbols := map[string]int{"I": 1, "V": 5, "X": 10, "L": 50, "C": 100, "D": 500, "M": 1000}
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ret := 0
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l := len(s)
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for i := 0; i < l-1; i++ {
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if symbols[s[i:i+1]] < symbols[s[i+1:i+2]] {
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ret -= symbols[s[i:i+1]]
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} else {
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ret += symbols[s[i:i+1]]
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}
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}
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ret += symbols[s[l-1:]]
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return ret
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}
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func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
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if l1 == nil {
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return l2
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}
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if l2 == nil {
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return l1
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}
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if l1.Val < l2.Val {
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l1.Next = mergeTwoLists(l1.Next, l2)
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return l1
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} else {
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l2.Next = mergeTwoLists(l1, l2.Next)
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return l2
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}
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}
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func removeElement(nums []int, val int) int {
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if len(nums) == 0 {
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return 0
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}
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i := 0
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for j := 0; j < len(nums); j++ {
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if nums[j] != val {
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nums[i] = nums[j]
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i++
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}
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}
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return i
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}
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func strStr(haystack string, needle string) int {
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switch {
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case len(needle) == 0:
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return 0
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case len(needle) > len(haystack):
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return -1
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case len(needle) == len(haystack):
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if needle == haystack {
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return 0
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}
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return -1
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}
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cursor := 0
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for i := 0; i < len(haystack); i++ {
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if haystack[i] == needle[cursor] {
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cursor++
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if cursor == len(needle) {
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return i - cursor + 1
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}
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} else {
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i -= cursor
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cursor = 0
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}
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}
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return -1
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}
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func searchInsert(nums []int, target int) int {
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left, right := 0, len(nums)
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for left < right {
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mid := (left + right) >> 1
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if nums[mid] >= target {
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right = mid
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} else {
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left = mid + 1
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}
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}
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return left
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}
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## 报数
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### 题目描述
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报数序列是一个整数序列,按照其中的整数的顺序进行报数,得到下一个数。其前五项如下:
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```
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1. 1
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2. 11
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3. 21
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4. 1211
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5. 111221
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```
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`1` 被读作 `"one 1"` (`"一个一"`) , 即 `11`。
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`11` 被读作 `"two 1s"` (`"两个一"`), 即 `21`。
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`21` 被读作 `"one 2"`, `"one 1"` (`"一个二"` , `"一个一"`) , 即 `1211`。
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给定一个正整数 n(1 ≤ n ≤ 30),输出报数序列的第 n 项。
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注意:整数顺序将表示为一个字符串。
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示例 1:
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```
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输入: 1
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输出: "1"
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```
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示例 2:
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```
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输入: 4
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输出: "1211"
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```
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func countAndSay(n int) string {
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buf := bytes.NewBufferString("1")
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for i := 2; i <= n; i++ {
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s := buf.String()
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c, l := s[0:1], len(s)
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buf.Reset()
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count := 0
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for j := 0; j < l; j++ {
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if c == s[j:j+1] {
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count++
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} else {
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buf.WriteByte(byte(48 + count))
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buf.WriteString(c)
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count = 1
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c = s[j : j+1]
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}
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}
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buf.WriteByte(byte(48 + count))
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buf.WriteString(c)
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}
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return buf.String()
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}
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func maxSubArray(nums []int) int {
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ans := nums[0]
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sum := 0
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for _, n := range nums {
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if sum > 0 {
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sum += n
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} else {
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sum = n
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}
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if sum > ans {
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ans = sum
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}
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}
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return ans
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}
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## 最后一个单词的长度
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### 题目描述
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给定一个仅包含大小写字母和空格 `' '` 的字符串,返回其最后一个单词的长度。
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如果不存在最后一个单词,请返回 `0` 。
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说明:一个单词是指由字母组成,但不包含任何空格的字符串。
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示例:
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```
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输入: "Hello World"
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输出: 5
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```
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func lengthOfLastWord(s string) int {
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if len(s) == 0 {
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return 0
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}
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space := []byte(" ")[0]
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for len(s) != 0 && s[len(s)-1] == space {
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s = s[:len(s)-1]
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}
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ret := 0
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for i := len(s) - 1; i >= 0; i-- {
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if s[i] != space {
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ret++
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} else {
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return ret
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}
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}
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return ret
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}

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