leetcode

Author: openinx

leetcode/README.md

@@ -525,5 +525,3 @@ h[n] = 2 * sum( h[i] * h[n-i] ) (1<= i < n )
 * [Minimum Window Substring](https://oj.leetcode.com/problems/minimum-window-substring/)  字符串滑动窗口， 维护4个变量： left 左端点 ； right 右端点 ; cur[256] 各字符当前窗口内的计数值； curlen 当前窗口内目标串字符的个数。 时间复杂度O(N). 类似的题： [Substring with Concatenation of All Words ](https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/), [Longest Substring Without Repeating Characters ](https://oj.leetcode.com/problems/longest-substring-without-repeating-characters/) . 
 
 
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leetcode/a.out

@@ -0,0 +1,70 @@
+#include <iostream>
+#include <algorithm>
+#include <string>
+#include <cstring>
+
+using namespace std;
+
+
+class Solution {
+public:
+
+	bool nativeInt(const char *s, int l ,int r){
+		if(l > r) return false;
+		for(; l <= r ; ++ l)
+			if(!isdigit(s[l])) return false;
+		return true;
+	}
+
+	bool isInt(const char *s, int l, int r){
+		if(l > r) return false;
+		if(s[l] == '-' || s[l] == '+') ++l;
+		if(l > r) return false;
+		return nativeInt(s, l , r);
+	}
+
+	bool isDecimal(const char *s, int l, int r){
+		int i, dot = -1;
+		if(l > r) return false;
+		if(s[l] == '-' || s[l] == '+') ++l;
+		for(i = l ; i <= r ; ++ i) if(s[i] == '-' || s[i] == '+') return false;
+		if(l > r) return false;
+		for(i = l ; i <= r ; ++ i) if(s[i] == '.') { dot = i ; break; }
+		if(dot == -1) return nativeInt(s, l , r);
+		if(dot == l) return nativeInt(s, l+1, r);
+		if(dot == r) return nativeInt(s, l, r-1);
+		return isInt(s, l, dot-1) && isInt(s, dot+1, r);
+	}
+
+    bool isNumber(const char *s) {
+    	int i , l, r, n = strlen(s), e = -1;
+    	l = 0 ; r = n - 1 ;
+    	while(l <= r && s[l] == ' ') ++l;
+    	while(l <= r && s[r] == ' ') --r;
+    	if(l > r ) return false;
+    	for(i = l ;i <= r ; ++ i) if(s[i] == 'e') { e = i;  break; }
+    	if(e == -1){
+    		return isDecimal(s, l, r);
+    	}
+    	return isDecimal(s, l, e-1) && isInt(s, e+1, r);
+    }
+};
+
+
+int main(){
+	char s[][100] = { 
+				  "0",
+				  " -0.1 ",
+				  " +0.1 ",
+				  " ++0.1 ",
+				  "abc",
+				  "1 a",
+				  "2e10",
+				  "2e-10",
+				  "2ee-10",
+				};
+	Solution sol;
+	for(int i = 0 ; i < 9 ; ++ i)
+		cout << sol.isNumber(s[i]) << endl;
+	return 0 ;
+}