Leetcode 2. Add Two Numbers

C++

class Solution {
public:
    vector<int> diStringMatch(string s) {
        int n = s.length(), lo = 0, hi = n;
        vector<int> perm(n + 1);
        for (int i = 0; i < n; ++i) {
            perm[i] = s[i] == 'I' ? lo++ : hi--;
        }
        perm[n] = lo; // 最后剩下一个数，此时 lo == hi
        return perm;
    }
};

Ruby

# @param {String} s
# @return {Integer[]}
def di_string_match(s)
    lo = 0
    hi = n = s.length

    perm = Array.new(hi+1)

    for i in 0...n
        ch = s[i]
        if ch == 'I'
            perm[i] = lo
            lo = lo + 1
        else
            perm[i] = hi
            hi = hi - 1
        end
    end
    perm[n] = lo
    return perm
end

Python

class Solution:
    def diStringMatch(self, s: str) -> List[int]:
        lo = 0
        hi = n = len(s)
        perm = [0] * (n+1)

        for i, ch in enumerate(s):
            if ch == 'I':
                perm[i] = lo
                lo = lo + 1
            else:
                perm[i] = hi
                hi = hi - 1

        perm[n] = lo
        return perm

Swift

class Solution {
    func diStringMatch(_ s: String) -> [Int] {
        let n = s.count
        var lo = 0, hi = n
        var perm: [Int] = Array<Int>(repeating: 0, count: n+1)

        for (i, ch) in s.enumerated() {
            if ch == "I" {
                perm[i] = lo
                lo = lo + 1
            } else {
                perm[i] = hi
                hi = hi - 1
            }
        }
        perm[n] = lo
        return perm
    }
}

标签：ch,return,lo,perm,Add,hi,Numbers,diStringMatch,Leetcode
来源： https://www.cnblogs.com/loganxu/p/16247968.html