算法细节系列（34）：再见字符串（2）
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算法细节系列（34）：再见字符串（2）
详细代码可以fork下Github上leetcode项目，不定期更新。
题目摘自leetcode：
- Leetcode 071. Simplify Path
- Leetcode 468. Validate IP Address
- Leetcode 165. Compare Version Numbers
- Leetcode 068. Text Justification
- Leetcode 151. Reverse Words in a String
Leetcode 071. Simplify Path
思路：
主要遇到三种符号
a. ".."
b. "."
c. "dir"
第一种情况模拟返回
第二种情况无作为
第三种情况进入文件夹下
主要是遇到".."，所以我们用stack来模拟这种返回上一层的操作代码如下：
public String simplifyPath(String path) {
Deque<String> stack = new ArrayDeque<>();
char[] cs = path.toCharArray();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < path.length(); ++i){
if (cs[i] == '/'){
if (i + 1 < path.length() && cs[i+1] == '/') i++;
String dir = sb.toString();
if (dir.isEmpty()) continue;
if (dir.equals("..")){
if (!stack.isEmpty()) stack.pop();
}
else if (dir.equals(".")){
}
else{
stack.push(dir);
}
sb = new StringBuilder();
}
else{
sb.append(cs[i]);
}
}
if (sb.length() != 0){
String dir = sb.toString();
if (dir.equals("..")){
if (!stack.isEmpty()) stack.pop();
}
else if (dir.equals(".")){
}
else{
stack.push(dir);
}
}
sb = new StringBuilder();
while (!stack.isEmpty()){
sb.append("/" + stack.pollLast());
}
return sb.toString().isEmpty() ? "/" : sb.toString();
}上面这代码够丑的，优化下，我用前一个”/”和下一个”/”作为读取数据的标志，所以在for循环结束还需要多判断一次。
我们可以用java自带的split方法进行简化：
public String simplifyPath(String path) {
Deque<String> stack = new LinkedList<>();
Set<String> omit = new HashSet<>(Arrays.asList("..",".",""));
for (String dir : path.split("/")){
if (dir.equals("..") && !stack.isEmpty()) stack.pop();
else if (!omit.contains(dir)) stack.push(dir);
}
StringBuilder sb = new StringBuilder();
while (!stack.isEmpty()) sb.append("/" + stack.pollLast());
return sb.length() == 0 ? "/" : sb.toString();
}Leetcode 468. Validate IP Address
找规则解析就好了，corner case比较多，要有耐心，代码如下：
String IPv4 = "IPv4";
String IPv6 = "IPv6";
public String validIPAddress(String IP) {
if (IP.isEmpty()) return "Neither";
String ans = IP.contains(".") ? IPv4 : IPv6;
if (ans.equals(IPv4)){
ans = validIPv4(IP) ? ans : "Neither";
}
else{
ans = validIPv6(IP) ? ans : "Neither";
}
return ans;
}
private boolean validIPv4(String IP){
if (IP.charAt(IP.length()-1) == '.' || IP.charAt(0) == '.') return false;
String[] nums = IP.split("\\.");
if (nums.length != 4) return false;
for (String num : nums){
char[] c = num.toCharArray();
if (num.isEmpty() || (c[0] == '0' && c.length != 1)) return false;
int n = 0;
int i = 0;
while (i < c.length && c[i] >= '0' && c[i] <= '9'){
n = 10 * n + c[i++] - '0';
}
if (!(n >= 0 && n <= 255) || i < c.length) return false;
}
return true;
}
private boolean validIPv6(String IP){
if (IP.charAt(IP.length()-1) == ':' || IP.charAt(0) == ':') return false;
String[] nums = IP.split(":");
if (nums.length != 8) return false;
for (String num : nums){
if (num.isEmpty()) return false;
char[] c = num.toCharArray();
if (c.length >= 5) return false;
for (int i = 0; i < c.length; ++i){
if (! (c[i] >= '0' && c[i] <= '9' || c[i] >= 'a' && c[i] <= 'f' || c[i] >= 'A' && c[i] <= 'F'))
return false;
}
}
return true;
}Leetcode 165. Compare Version Numbers
版本的比较，有点类似基数排序，很简单，先split(“.”)，遇到大小相等的版本号（同一级别下），进入下一级别的比较。
代码如下：
public int compareVersion(String version1, String version2) {
String[] v1 = version1.split("\\.");
String[] v2 = version2.split("\\.");
int min = Math.min(v1.length, v2.length);
int i = 0;
while (i < min && Integer.parseInt(v1[i]) == Integer.parseInt(v2[i])) i++;
if (i == min){
if (v1.length > v2.length){
for (int j = i; j < v1.length; ++j){
if (Integer.parseInt(v1[j]) != 0) return 1;
}
}
else if (v1.length < v2.length){
for (int j = i; j < v2.length; ++j){
if (Integer.parseInt(v2[j]) != 0) return -1;
}
}
return 0;
}
return Integer.parseInt(v1[i]) > Integer.parseInt(v2[i]) ? 1 : -1;
}代码简化思路：
version1 = "1.0.1.3.4.5"
version2 = "1.0"
可以看成：
version2 = "1.0.0.0.0.0"代码如下：
public int compareVersion(String version1, String version2) {
String[] v1 = version1.split("\\.");
String[] v2 = version2.split("\\.");
int max = Math.max(v1.length, v2.length);
for (int i = 0; i < max; ++i){
int num1 = i < v1.length ? Integer.parseInt(v1[i]) : 0;
int num2 = i < v2.length ? Integer.parseInt(v2[i]) : 0;
if (num1 < num2) return -1;
if (num1 > num2) return 1;
}
return 0;
}Leetcode 068. Text Justification
贪心算法，比较细节，代码逻辑较复杂。
代码如下：（第一版）
public List<String> fullJustify(String[] words, int maxWidth) {
List<String> ans = new ArrayList<>();
if (words.length == 0 || maxWidth == 0){
ans.add("");
return ans;
}
String res = greedy(words, maxWidth);
while (!res.isEmpty()){
ans.add(res);
res = greedy(words, maxWidth);
}
return ans;
}
int pos = 0;
private String greedy(String[] words, int maxWidth){
if (pos == words.length) return "";
String ans = words[pos];
int cnt = ans.length();
for (int i = pos + 1; i < words.length; ++i){
if (cnt + words[i].length() + 1 <= maxWidth){
ans += " " + words[i];
cnt = ans.length();
pos = i + 1;
}
else{
pos = i;
int space = maxWidth - cnt;
String[] all = ans.split(" ");
int len = all.length - 1;
StringBuilder sb = new StringBuilder();
if (len == 0){
sb.append(all[0]);
for (int k = 0; k < space; ++k){
sb.append(" ");
}
return sb.toString();
}
if (space % len == 0){
for (int k = 0; k < all.length; ++k){
sb.append(all[k] + " ");
for (int l = 0; l < space / len; ++l){
sb.append(" ");
}
}
return sb.toString().trim();
}
else{
int re = space % len;
int sp = space / len;
for (int k = 0; k < all.length; ++k){
sb.append(all[k] + " " + ((re != 0) ? " " : ""));
re = Math.max(--re,0);
for (int l = 0; l < sp; ++l){
sb.append(" ");
}
}
return sb.toString().trim();
}
}
}
pos = words.length;
int space = maxWidth - cnt;
String[] all = ans.split(" ");
int len = all.length - 1;
StringBuilder sb = new StringBuilder();
if (len == 0){
sb.append(all[0]);
for (int k = 0; k < space; ++k){
sb.append(" ");
}
return sb.toString();
}
sb = new StringBuilder(ans);
for (int k = 0; k < space; ++k){
sb.append(" ");
}
return sb.toString();
}for循环可以用while替代，一次找到需要组合的words，这样for中else的逻辑和for循环外的逻辑可以合并。
代码如下：（第二版本）
int pos = 0;
private String greedy(String[] words, int maxWidth){
if (pos == words.length) return "";
int cnt = words[pos].length();
int i = pos + 1;
while (i < words.length && cnt + words[i].length() + 1 <= maxWidth){
cnt += words[i++].length() + 1;
}
int space = maxWidth - cnt;
int len = i - pos - 1;
if (len == 0 || i == words.length){
StringBuilder sb = new StringBuilder();
for (int k = pos; k < i; ++k){
sb.append(words[k] + " ");
}
sb.deleteCharAt(sb.length()-1);
for (int k = 0; k < space; ++k){
sb.append(" ");
}
pos = i;
return sb.toString();
}
StringBuilder sb = new StringBuilder();
if (space % len == 0){
for (int k = pos; k < i; ++k){
sb.append(words[k] + " ");
for (int l = 0; l < space / len; ++l){
sb.append(" ");
}
}
pos = i;
return sb.toString().trim();
}
else{
int re = space % len;
int sp = space / len;
for (int k = pos; k < i; ++k){
sb.append(words[k] + " " + ((re != 0) ? " " : ""));
re = Math.max(--re,0);
for (int l = 0; l < sp; ++l){
sb.append(" ");
}
}
pos = i;
return sb.toString().trim();
}
}单独开一个greedy方法也没谁了，还定义了全局变量pos，与其这样，不如放在一个方法内。
代码如下：（第三版本）
public List<String> fullJustify(String[] words, int maxWidth) {
List<String> ans = new ArrayList<>();
int pos = 0;
while (pos < words.length){
int cnt = words[pos].length();
int i = pos + 1;
while (i < words.length && cnt + words[i].length() + 1 <= maxWidth){
cnt += words[i++].length() + 1;
}
int space = maxWidth - cnt;
int len = i - pos - 1;
StringBuilder sb = new StringBuilder();
if (len == 0 || i == words.length){
sb = new StringBuilder();
for (int k = pos; k < i; ++k){
sb.append(words[k] + " ");
}
sb.deleteCharAt(sb.length()-1);
for (int k = 0; k < space; ++k){
sb.append(" ");
}
pos = i;
ans.add(sb.toString());
}
else{
sb = new StringBuilder();
if (space % len == 0){
for (int k = pos; k < i; ++k){
sb.append(words[k] + " ");
for (int l = 0; l < space / len; ++l){
sb.append(" ");
}
}
pos = i;
ans.add(sb.toString().trim());
}
else{
int re = space % len;
int sp = space / len;
for (int k = pos; k < i; ++k){
sb.append(words[k] + " " + ((re != 0) ? " " : ""));
re = Math.max(--re,0);
for (int l = 0; l < sp; ++l){
sb.append(" ");
}
}
pos = i;
ans.add(sb.toString().trim());
}
}
}
return ans;
}其实，space % len 都不需要单独出逻辑来，因为如果re为0就不append，如果re不为0，把这部分加上去就好了。
代码如下：（第四版本）
public List<String> fullJustify(String[] words, int maxWidth) {
List<String> ans = new ArrayList<>();
int pos = 0;
while (pos < words.length){
int cnt = words[pos].length();
int i = pos + 1;
while (i < words.length && cnt + words[i].length() + 1 <= maxWidth){
cnt += words[i++].length() + 1;
}
int space = maxWidth - cnt;
int len = i - pos - 1;
StringBuilder sb = new StringBuilder();
if (len == 0 || i == words.length){
sb = new StringBuilder();
for (int k = pos; k < i; ++k){
sb.append(words[k] + " ");
}
sb.deleteCharAt(sb.length()-1);
for (int k = 0; k < space; ++k){
sb.append(" ");
}
ans.add(sb.toString());
}
else{
sb = new StringBuilder();
int re = space % len;
int sp = space / len;
for (int k = pos; k < i; ++k){
sb.append(words[k] + " " + ((re != 0) ? " " : ""));
re = Math.max(--re,0);
for (int l = 0; l < sp; ++l){
sb.append(" ");
}
}
ans.add(sb.toString().trim());
}
pos = i;
}
return ans;
}比较满意了。
Leetcode 151. Reverse Words in a String
用自带的java库函数做一把。
public String reverseWords(String s) {
String[] words = s.trim().split(" +");
Stack<String> stack = new Stack<>();
for (String word : words){
stack.push(word);
}
StringBuilder sb = new StringBuilder();
while (!stack.isEmpty()){
sb.append(stack.pop() + " ");
}
return sb.toString().trim();
}三行版本：
public String reverseWords(String s) {
String[] words = s.trim().split(" +");
Collections.reverse(Arrays.asList(words));
return String.join(" ", words);
}本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2017年06月13日，如有侵权请联系 cloudcommunity@tencent.com 删除
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