5.2 Nesting

Determine whether given string of parentheses is properly nested.
需要注意空间复杂度为1，不能使用stack。
A string S consisting of N characters is called properly nested if:
S is empty;
S has the form “(U)” where U is a properly nested string;
S has the form “VW” where V and W are properly nested strings.
For example, string “(()(())())” is properly nested but string “())” isn’t.
Write a function:
class Solution { public int solution(String S); }
that, given a string S consisting of N characters, returns 1 if string S is properly nested and 0 otherwise.
For example, given S = “(()(())())”, the function should return 1 and given S = “())”, the function should return 0, as explained above.
Assume that:
N is an integer within the range [0..1,000,000];
string S consists only of the characters “(” and/or “)”.
Complexity:
expected worst-case time complexity is O(N);
expected worst-case space complexity is O(1) (not counting the storage required for input arguments).

Solution

class Solution {    public int solution(String S) {        // write your code in Java SE 8        int count = 0;        // '(' -> 1   ')' -> -1        for(int i=0; i<S.length(); i++){            if(S.charAt(i)=='(') count++;            else count--;            if(count<0){                return 0;            }           }        if(count==0) return 1;        else return 0;    }}

so easy. 这种方法只适用于一种符号的情况。如果存在（），[]，需要保存两种符号出现的顺序，[(])非法