README

Author: kimi0230

Leetcode/0003.Longest-Substring-Without-Repeating-Characters/README.md

@@ -43,4 +43,104 @@ O(n)
 
 ## 來源
 * https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0003.Longest-Substring-Without-Repeating-Characters/
-* https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
+* https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
+
+## 解答
+https://github.com/kimi0230/LeetcodeGolang/blob/master/Leetcode/0003.Longest-Substring-Without-Repeating-Characters/Longest-Substring-Without-Repeating-Characters.go
+
+```go
+package longestSubstringwithoutrepeatingcharacters
+
+// LengthOfLongestSubstring 暴力解
+func LengthOfLongestSubstring(s string) int {
+	slength := len(s)
+	if slength == 0 || slength == 1 {
+		return slength
+	}
+
+	tmpLen := 1
+	var maxLen = 1
+
+	for i := 1; i < slength; i++ {
+		// 往前找前幾個視窗
+		j := i - tmpLen
+
+		for ; j < i; j++ {
+			if s[j] == s[i] { // 如果相同，那麼和S[J]到S[I-1]中間的肯定不相同，所以可以直接計算得到
+				tmpLen = i - j
+				break
+			}
+		}
+
+		if j == i { // 都不相同
+			tmpLen++
+		}
+
+		if tmpLen > maxLen {
+			maxLen = tmpLen
+		}
+	}
+
+	return maxLen
+}
+
+// LengthOfLongestSubstringMap 用map 紀錄是否重複.
+func LengthOfLongestSubstringMap(s string) int {
+	slength := len(s)
+	if slength == 0 || slength == 1 {
+		return slength
+	}
+
+	charMap := make(map[byte]bool)
+	maxLen, left, right := 0, 0, 0
+	for i := 0; i < slength; i++ {
+		if ok := charMap[s[i]]; ok {
+			// 有找到
+			charMap[s[left]] = false
+			left++
+		} else {
+			charMap[s[i]] = true
+			right++
+		}
+
+		if maxLen < right-left {
+			maxLen = right - left
+		}
+		if left+maxLen >= slength || right >= len(s) {
+			break
+		}
+	}
+	return maxLen
+}
+
+// LengthOfLongestSubstringBit 用map效能不好時可使用數組改善
+func LengthOfLongestSubstringBit(s string) int {
+	slength := len(s)
+	if slength == 0 || slength == 1 {
+		return slength
+	}
+
+	// ASCII 0~255
+	charMap := [256]bool{}
+	maxLen, left, right := 0, 0, 0
+	for i := 0; i < slength; i++ {
+		if ok := charMap[s[i]]; ok {
+			// 有找到
+			charMap[s[left]] = false
+			left++
+		} else {
+			charMap[s[i]] = true
+			right++
+		}
+
+		if maxLen < right-left {
+			maxLen = right - left
+		}
+		if left+maxLen >= slength || right >= len(s) {
+			break
+		}
+	}
+	return maxLen
+}
+
+```

Leetcode/0015.3Sum/README.md

@@ -32,4 +32,91 @@ A solution set is:
 
 ## 來源
 * https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0015.3Sum/
-* https://leetcode-cn.com/problems/3sum/
+* https://leetcode-cn.com/problems/3sum/
+
+
+## 解答
+https://github.com/kimi0230/LeetcodeGolang/blob/master/Leetcode/0015.3Sum/3Sum.go
+
+```go
+package threesum
+
+import (
+	"sort"
+)
+
+// ThreeSumBurst : 暴力解 : O(n^3)
+func ThreeSumBurst(nums []int) [][]int {
+	result := [][]int{}
+	sort.Ints(nums) // O(n log n)
+	for i := 0; i < len(nums); i++ {
+		// 需要跟上一次不同
+		if i > 0 && nums[i] == nums[i-1] {
+			continue
+		}
+		for j := i + 1; j < len(nums); j++ {
+			// 需要跟上一次不同
+			if j > i+1 && nums[j] == nums[j-1] {
+				continue
+			}
+			for k := j + 1; k < len(nums); k++ {
+				if nums[i]+nums[j]+nums[k] == 0 {
+					result = append(result, []int{nums[i], nums[j], nums[k]})
+				}
+			}
+		}
+	}
+	return result
+}
+
+/*
+ThreeSumDoublePoint : 最佳解, 排序 + 雙指針法 (滑動視窗) O(n^2)
+1. 特判，對於數組長度 n，如果數組為 null 或者數組長度小於 3，返回 []。
+2. 對數組進行排序。
+3. 遍歷排序後數組：
+	* 對於重複元素：跳過，避免出現重複解
+	* 令左指針 L=i+1，右指針 R=n−1，當 L<R 時，執行循環：
+		* 當nums[i]+nums[L]+nums[R]==0，執行循環，判斷左界和右界是否和下一位置重複，
+		去除重複解。並同時將 L,R 移到下一位置，尋找新的解
+		* 若和大於 0，說明 nums[R] 太大，R 左移
+		* 若和小於 0，說明 nnums[L] 太小，L 右移
+*/
+func ThreeSumDoublePoint(nums []int) [][]int {
+	result := [][]int{}
+	if len(nums) < 3 {
+		return result
+	}
+	sort.Ints(nums) // O(n log n)
+	start, end, addNum := 0, 0, 0
+	for i := 1; i < len(nums)-1; i++ {
+		start, end = 0, len(nums)-1
+		if i > 1 && nums[i] == nums[i-1] {
+			// 去掉重複
+			start = i - 1
+		}
+		for start < i && end > i {
+			if start > 0 && nums[start] == nums[start-1] {
+				// 去掉重複
+				start++
+				continue
+			}
+			if end < (len(nums)-1) && nums[end] == nums[end+1] {
+				// 去掉重複
+				end--
+				continue
+			}
+			addNum = nums[start] + nums[end] + nums[i]
+			if addNum == 0 {
+				result = append(result, []int{nums[start], nums[i], nums[end]})
+				start++
+				end--
+			} else if addNum > 0 {
+				end--
+			} else {
+				start++
+			}
+		}
+	}
+	return result
+}
+```

Leetcode/0027.Remove-Element/README.md

@@ -62,4 +62,49 @@ for (int i = 0; i < len; i++) {
 ## 來源
 * https://books.halfrost.com/leetcode/ChapterFour/0001~0099/0027.Remove-Element/
 * https://leetcode-cn.com/problems/remove-element/
-* https://mp.weixin.qq.com/s/wj0T-Xs88_FHJFwayElQlA
+* https://mp.weixin.qq.com/s/wj0T-Xs88_FHJFwayElQlA
+
+## 解答
+https://github.com/kimi0230/LeetcodeGolang/blob/master/Leetcode/0027.Remove-Element/Remove-Element.go
+
+```go
+package removeelement
+
+/*
+雙指針法
+雙指針法（快慢指針法）在數組和鍊錶的操作中是非常常見的，很多考察數組和鍊錶操作的面試題，都使用雙指針法
+*/
+func RemoveElementDoublePoint(nums []int, val int) int {
+	if len(nums) <= 0 {
+		return 0
+	}
+	slowIndex := 0
+	for fastIndex := 0; fastIndex < len(nums); fastIndex++ {
+		if nums[fastIndex] != val {
+			if fastIndex != slowIndex {
+				nums[fastIndex], nums[slowIndex] = nums[slowIndex], nums[fastIndex]
+			}
+			slowIndex++
+		}
+	}
+	return slowIndex
+}
+
+/*
+RemoveElement :
+*/
+func RemoveElement(nums []int, val int) int {
+	size := len(nums)
+	i := 0
+	for i < size {
+		if nums[i] == val {
+			nums = append(nums[:i], nums[i+1:]...)
+			size--
+			// fmt.Println(nums)
+		} else {
+			i++
+		}
+	}
+	return len(nums)
+}
+```