『LeetCode』1662 检查两个字符串数组是否相等

题目

给你两个字符串数组 word1 和 word2 。如果两个数组表示的字符串相同，返回true；否则，返回 false

数组表示的字符串 是由数组中的所有元素 按顺序 连接形成的字符串。

示例 1：

输入：word1 = ["ab", "c"], word2 = ["a", "bc"]
输出：true
解释：
word1 表示的字符串为 "ab" + "c" -> "abc"
word2 表示的字符串为 "a" + "bc" -> "abc"
两个字符串相同，返回 true

示例 2：

输入：word1 = ["a", "cb"], word2 = ["ab", "c"]
输出：false

示例 3：

输入：word1 = ["abc", "d", "defg"], word2 = ["abcddefg"]
输出：true

提示：

1 <= word1.length, word2.length <= 10^{3}
1 <= word1[i].length, word2[i].length <= 10^{3}
1 <= sum(word1[i].length), sum(word2[i].length) <= 10^{3}
word1[i] 和 word2[i] 由小写字母组成

题解

【检查两个字符串数组是否相等】简单一行模拟

模拟

最简单的方法就是拼接然后模拟，一行代码即可。但空间复杂度为 \(O(n)\)

更好一些的方法是用指针滑动判断对比。

# Code language: Python
# 滑动指针（迭代器）
class Solution:
    def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
        return all(a == b for a, b in zip_longest(chain(*word1), chain(*word2)))

# Code language: Python
# 拼接
class Solution:
    def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:
        return "".join(word1) == "".join(word2)

// Code language: Java
class Solution {
    public boolean arrayStringsAreEqual(String[] word1, String[] word2) {
        int n1 = word1.length, n2 = word2.length, i = 0, j = 0, p = 0, q = 0;
        while (i < n1 && j < n2) {
            if (word1[i].charAt(p) != word2[j].charAt(q)) return false;
            if (++p >= word1[i].length()) {
                ++i; p = 0;
            }
            if (++q >= word2[j].length()) {
                ++j; q = 0;
            }
        }
        return i == n1 && j == n2 && p == 0 && q == 0;
    }
}

// Code language: C++
class Solution {
public:
    bool arrayStringsAreEqual(vector<string>& word1, vector<string>& word2) {
        int n1 = word1.size(), n2 = word2.size(), i = 0, j = 0, p = 0, q = 0;
        while (i < n1 && j < n2) {
            if (word1[i][p] != word2[j][q]) return false;
            if (++p >= word1[i].size()) ++i, p = 0;
            if (++q >= word2[j].size()) ++j, q = 0;
        }
        return i == n1 && j == n2 && p == 0 && q == 0;
    }
};

/* Code language: JavaScript */
/**
 * @param {string[]} word1
 * @param {string[]} word2
 * @return {boolean}
 */
var arrayStringsAreEqual = function(word1, word2) {
    let n1 = word1.length, n2 = word2.length, i = 0, j = 0, p = 0, q = 0;
    while (i < n1 && j < n2) {
        if (word1[i][p] != word2[j][q]) return false;
        if (++p >= word1[i].length) ++i, p = 0;
        if (++q >= word2[j].length) ++j, q = 0;
    }
    return i == n1 && j == n2 && p == 0 && q == 0;
};

/* Code language: Go */
func arrayStringsAreEqual(word1 []string, word2 []string) bool {
    n1, n2 := len(word1), len(word2)
    i, j, p, q := 0, 0, 0, 0
    for i < n1 && j < n2 {
        if word1[i][p] != word2[j][q] {
            return false
        }
        if p++; p >= len(word1[i]) {
            i++; p = 0;
        }
        if q++; q >= len(word2[j]) {
            j++; q = 0;
        }
    }
    return i == n1 && j == n2 && p == 0 && q == 0
}

时间复杂度: \(O(n)\)
空间复杂度: \(O(1)\)

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『LeetCode』1662 检查两个字符串数组是否相等

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题解

模拟