finish Problem 32

我的答案，不是最快的，但我觉得是最清晰的思路

Author: aQua

Algorithms/0032.longest-valid-parentheses/README.md

@@ -8,8 +8,20 @@ For `"(()"`, the longest valid parentheses substring is `"()"`, which has length
 Another example is `")()())"`, where the longest valid parentheses substring is `"()()"`, which has length = 4.
 
 ## 解题思路
-
-
+1.记录每个符号的状态，`(`一律对应于`0`；`)`如果能够和前面的配上对，就记录为`2`，否则，记录为`0`
+```
+) ( ( ) ( ) ) ) ( ( ( ( ( ) ) ) ) (
+```
+形成记录
+```
+0 0 0 2 0 2 2 0 0 0 0 0 0 2 2 2 2 0
+```
+
+2.从左往右检查record，如果record[i]==2，record[j]==0，且record[j+1:i]中没有0，则record[i]=1,record[j]=1，那么，上面的record就变成了
+```
+0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0
+```
+3.统计record中，最多的连续为1的次数，就是结果。
 ## 总结
 
-
+根据记录统计，往往是最清晰的

Algorithms/0032.longest-valid-parentheses/longest-valid-parentheses.go

@@ -1,31 +1,38 @@
 package Problem0032
 
 func longestValidParentheses(s string) int {
-	var i, j, max, temp int
+	var left, max, temp int
 	record := make([]int, len(s))
-	lens := len(s)
 
-	for i < lens {
-		for i < lens-1 && !(s[i] == '(' && s[i+1] == ')') {
-			i++
+	// 统计Record
+	for i, b := range s {
+		if b == '(' {
+			left++
+		} else if left > 0 {
+			left--
+			record[i] = 2
 		}
-		j = i + 1
-		for i >= 0 && s[i] == '(' && j < lens && s[j] == ')' {
+	}
+
+	// 修改record
+	for i := 0; i < len(record); i++ {
+		if record[i] == 2 {
+			j := i - 1
+			for record[j] != 0 {
+				j--
+			}
 			record[i], record[j] = 1, 1
-			i--
-			j++
 		}
-
-		i = j
 	}
 
+	// 统计结果
 	for _, r := range record {
 		if r == 0 {
 			temp = 0
-		} else {
-			temp++
+			continue
 		}
 
+		temp++
 		if temp > max {
 			max = temp
 		}

Algorithms/0032.longest-valid-parentheses/longest-valid-parentheses_test.go

@@ -28,17 +28,21 @@ func Test_Problem0032(t *testing.T) {
 	ast := assert.New(t)
 
 	qs := []question{
-
 		question{
-			para{"(()())"},
-			ans{6},
+			para{")(()()))((((())))("},
+			ans{8},
 		},
 
 		question{
 			para{"()(()"},
 			ans{2},
 		},
 
+		question{
+			para{"(()())"},
+			ans{6},
+		},
+
 		question{
 			para{"(()"},
 			ans{2},