1111. The effective depth of nesting brackets

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Title Description

Thinking

class Solution {
    public int[] maxDepthAfterSplit(String seq) {
        if(seq == null){
            return new int[0];
        }
        int idx = 0;
        int[] res = new int[seq.length()];
        for(char c : seq.toCharArray()){
            res[idx++] = c == '(' ? (idx&1) : ((idx+1)&1);
        }
        return res;
    }
}

public int[] maxDepthAfterSplit(String seq) {
        int[] res = new int[seq.length()];
        int index = 0;
        for(char c : seq.toCharArray())
        {
            //如果当前字符是左括号
            if(c == '(')
            {
                //如果index++后是奇数，index在数组中的对应位置是0，否则是1 
                res[index++] = (index + 1) & 1 ;
            }
            //如果当前字符是右括号
            else{
                //如果index++后是奇数，index在数组中的对应位置是1，否则是0 
                res[index++] = index & 1 ;
            }
        }
        return res;
    }