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feat: add solutions to lc problem: No.1754
No.1754.Largest Merge Of Two Strings
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‎solution/1700-1799/1754.Largest Merge Of Two Strings/README.md

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<ul>
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<li>如果 <code>word1</code> 非空，将 <code>word1</code> 中的第一个字符附加到 <code>merge</code> 的末尾，并将其从 <code>word1</code> 中移除。
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<ul>
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<li>例如，<code>word1 = "abc" </code>且 <code>merge = "dv"</code> ，在执行此选项操作之后，<code>word1 = "bc"</code> ，同时 <code>merge = "dva"</code> 。</li>
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</ul>
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<li>例如，<code>word2 = "abc" </code>且 <code>merge = ""</code> ，在执行此选项操作之后，<code>word2 = "bc"</code> ，同时 <code>merge = "a"</code> 。</li>
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</ul>
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</li>
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</ul>
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<p>返回你可以构造的字典序 <strong>最大</strong> 的合并字符串<em> </em><code>merge</code><em> 。</em></p>
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 双指针**
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我们用指针 $i$ 和 $j$ 分别指向字符串 `word1` 和 `word2` 的第一个字符。然后循环，每次比较 $word1[i:]$ 和 $word2[j:]$ 的大小，如果 $word1[i:]$ 比 $word2[j:]$ 大，那么我们就将 $word1[i]$ 加入答案，否则我们就将 $word2[j]$ 加入答案。循环，直至 $i$ 到达字符串 `word1` 的末尾，或者 $j$ 到达字符串 `word2` 的末尾。
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然后我们将剩余的字符串加入答案即可。
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时间复杂度 $O(n^2)$。其中 $n$ 是字符串 `word1` 和 `word2` 的长度之和。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def largestMerge(self, word1: str, word2: str) -> str:
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i = j = 0
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ans = []
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while i < len(word1) and j < len(word2):
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if word1[i:] > word2[j:]:
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ans.append(word1[i])
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i += 1
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else:
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ans.append(word2[j])
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j += 1
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ans.append(word1[i:])
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ans.append(word2[j:])
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return "".join(ans)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String largestMerge(String word1, String word2) {
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int m = word1.length(), n = word2.length();
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int i = 0, j = 0;
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StringBuilder ans = new StringBuilder();
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while (i < m && j < n) {
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boolean gt = word1.substring(i).compareTo(word2.substring(j)) > 0;
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ans.append(gt ? word1.charAt(i++) : word2.charAt(j++));
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}
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ans.append(word1.substring(i));
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ans.append(word2.substring(j));
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return ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string largestMerge(string word1, string word2) {
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int m = word1.size(), n = word2.size();
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int i = 0, j = 0;
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string ans;
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while (i < m && j < n) {
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bool gt = word1.substr(i) > word2.substr(j);
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ans += gt ? word1[i++] : word2[j++];
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}
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ans += word1.substr(i);
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ans += word2.substr(j);
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return ans;
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}
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};
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```
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### **Go**
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```go
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func largestMerge(word1 string, word2 string) string {
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m, n := len(word1), len(word2)
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i, j := 0, 0
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var ans strings.Builder
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for i < m && j < n {
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if word1[i:] > word2[j:] {
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ans.WriteByte(word1[i])
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i++
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} else {
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ans.WriteByte(word2[j])
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j++
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}
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}
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ans.WriteString(word1[i:])
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ans.WriteString(word2[j:])
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return ans.String()
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}
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```
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### **...**

‎solution/1700-1799/1754.Largest Merge Of Two Strings/README_EN.md

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<ul>
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<li>If <code>word1</code> is non-empty, append the <strong>first</strong> character in <code>word1</code> to <code>merge</code> and delete it from <code>word1</code>.
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<ul>
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<li>For example, if <code>word1 = &quot;abc&quot; </code>and <code>merge = &quot;dv&quot;</code>, then after choosing this operation, <code>word1 = &quot;bc&quot;</code> and <code>merge = &quot;dva&quot;</code>.</li>
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</ul>
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<li>For example, if <code>word2 = &quot;abc&quot; </code>and <code>merge = &quot;&quot;</code>, then after choosing this operation, <code>word2 = &quot;bc&quot;</code> and <code>merge = &quot;a&quot;</code>.</li>
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</ul>
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</li>
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</ul>
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<p>Return <em>the lexicographically <strong>largest</strong> </em><code>merge</code><em> you can construct</em>.</p>
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### **Python3**
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```python
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class Solution:
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def largestMerge(self, word1: str, word2: str) -> str:
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i = j = 0
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ans = []
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while i < len(word1) and j < len(word2):
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if word1[i:] > word2[j:]:
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ans.append(word1[i])
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i += 1
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else:
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ans.append(word2[j])
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j += 1
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ans.append(word1[i:])
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ans.append(word2[j:])
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return "".join(ans)
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```
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### **Java**
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```java
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class Solution {
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public String largestMerge(String word1, String word2) {
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int m = word1.length(), n = word2.length();
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int i = 0, j = 0;
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StringBuilder ans = new StringBuilder();
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while (i < m && j < n) {
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boolean gt = word1.substring(i).compareTo(word2.substring(j)) > 0;
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ans.append(gt ? word1.charAt(i++) : word2.charAt(j++));
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}
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ans.append(word1.substring(i));
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ans.append(word2.substring(j));
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return ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string largestMerge(string word1, string word2) {
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int m = word1.size(), n = word2.size();
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int i = 0, j = 0;
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string ans;
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while (i < m && j < n) {
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bool gt = word1.substr(i) > word2.substr(j);
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ans += gt ? word1[i++] : word2[j++];
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}
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ans += word1.substr(i);
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ans += word2.substr(j);
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return ans;
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}
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};
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```
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### **Go**
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```go
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func largestMerge(word1 string, word2 string) string {
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m, n := len(word1), len(word2)
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i, j := 0, 0
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var ans strings.Builder
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for i < m && j < n {
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if word1[i:] > word2[j:] {
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ans.WriteByte(word1[i])
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i++
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} else {
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ans.WriteByte(word2[j])
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j++
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}
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}
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ans.WriteString(word1[i:])
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ans.WriteString(word2[j:])
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return ans.String()
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}
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```
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### **...**

‎solution/1700-1799/1754.Largest Merge Of Two Strings/Solution.cpp

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class Solution {
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public:
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string largestMerge(string word1, string word2) {
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int m = word1.size(), n = word2.size();
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int i = 0, j = 0;
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string ans;
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while (i < m && j < n) {
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bool gt = word1.substr(i) > word2.substr(j);
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ans += gt ? word1[i++] : word2[j++];
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}
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ans += word1.substr(i);
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ans += word2.substr(j);
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return ans;
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}
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};

‎solution/1700-1799/1754.Largest Merge Of Two Strings/Solution.go

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func largestMerge(word1 string, word2 string) string {
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m, n := len(word1), len(word2)
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i, j := 0, 0
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var ans strings.Builder
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for i < m && j < n {
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if word1[i:] > word2[j:] {
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ans.WriteByte(word1[i])
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i++
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} else {
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ans.WriteByte(word2[j])
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j++
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}
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}
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ans.WriteString(word1[i:])
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ans.WriteString(word2[j:])
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return ans.String()
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}

‎solution/1700-1799/1754.Largest Merge Of Two Strings/Solution.java

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class Solution {
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public String largestMerge(String word1, String word2) {
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int m = word1.length(), n = word2.length();
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int i = 0, j = 0;
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StringBuilder ans = new StringBuilder();
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while (i < m && j < n) {
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boolean gt = word1.substring(i).compareTo(word2.substring(j)) > 0;
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ans.append(gt ? word1.charAt(i++) : word2.charAt(j++));
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}
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ans.append(word1.substring(i));
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ans.append(word2.substring(j));
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return ans.toString();
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}
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}

‎solution/1700-1799/1754.Largest Merge Of Two Strings/Solution.py

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class Solution:
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def largestMerge(self, word1: str, word2: str) -> str:
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i = j = 0
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ans = []
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while i < len(word1) and j < len(word2):
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if word1[i:] > word2[j:]:
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ans.append(word1[i])
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i += 1
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else:
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ans.append(word2[j])
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j += 1
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ans.append(word1[i:])
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ans.append(word2[j:])
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return "".join(ans)

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