java 解法, 执行用时: 1 ms, 内存消耗: 41.2 MB, 提交时间: 2022-12-24 12:06:36

class Solution {
    public boolean oneEditAway(String first, String second) {
        int m = first.length(), n = second.length();
        if (n - m == 1) {
            return oneInsert(first, second);
        } else if (m - n == 1) {
            return oneInsert(second, first);
        } else if (m == n) {
            boolean foundDifference = false;
            for (int i = 0; i < m; i++) {
                if (first.charAt(i) != second.charAt(i)) {
                    if (!foundDifference) {
                        foundDifference = true;
                    } else {
                        return false;
                    }
                }
            }
            return true;
        } else {
            return false;
        }
    }

    public boolean oneInsert(String shorter, String longer) {
        int length1 = shorter.length(), length2 = longer.length();
        int index1 = 0, index2 = 0;
        while (index1 < length1 && index2 < length2) {
            if (shorter.charAt(index1) == longer.charAt(index2)) {
                index1++;
            }
            index2++;
            if (index2 - index1 > 1) {
                return false;
            }
        }
        return true;
    }
}

javascript 解法, 执行用时: 88 ms, 内存消耗: 43 MB, 提交时间: 2022-12-24 12:06:14

/**
 * @param {string} first
 * @param {string} second
 * @return {boolean}
 */
var oneEditAway = function(first, second) {
    const m = first.length, n = second.length;
    if (n - m === 1) {
        return oneInsert(first, second);
    } else if (m - n === 1) {
        return oneInsert(second, first);
    } else if (m === n) {
        let foundDifference = false;
        for (let i = 0; i < m; i++) {
            if (first[i] != second[i]) {
                if (!foundDifference) {
                    foundDifference = true;
                } else {
                    return false;
                }
            }
        }
        return true;
    } else {
        return false;
    }
}

const oneInsert = (shorter, longer) => {
    const length1 = shorter.length, length2 = longer.length;
    let index1 = 0, index2 = 0;
    while (index1 < length1 && index2 < length2) {
        if (shorter[index1] == longer[index2]) {
            index1++;
        }
        index2++;
        if (index2 - index1 > 1) {
            return false;
        }
    }
    return true;
};

golang 解法, 执行用时: 0 ms, 内存消耗: 2.3 MB, 提交时间: 2022-12-24 12:05:58

func oneEditAway(first, second string) bool {
    m, n := len(first), len(second)
    if m < n {
        return oneEditAway(second, first)
    }
    if m-n > 1 {
        return false
    }
    for i, ch := range second {
        if first[i] != byte(ch) {
            if m == n {
                return first[i+1:] == second[i+1:]
            }
            return first[i+1:] == second[i:]
        }
    }
    return true
}

python3 解法, 执行用时: 32 ms, 内存消耗: 14.9 MB, 提交时间: 2022-12-24 12:05:12

class Solution:
    def oneEditAway(self, first: str, second: str) -> bool:
        m, n = len(first), len(second)
        if m < n:
            return self.oneEditAway(second, first)
        if m - n > 1:
            return False
        for i, (x, y) in enumerate(zip(first, second)):
            if x != y:
                return first[i + 1:] == second[i + 1:] if m == n else first[i + 1:] == second[i:]  # 注：改用下标枚举可达到 O(1) 空间复杂度
        return True