标签：优先 ring clean 方法 tail pac color rds one
344. Reverse String
解法一（暴力法）：
直接从两头往中间走，同时交换两边的字符即可
首位对调位置。
class Solution { public void reverseString(char[] s) { int tail = s.length-1; for(int i = 0; i < s.length/2; i++){ char temp = s[i]; s[i] = s[tail-i]; s[tail-i] = temp; } } }
解法二：
利用swap函数求解
class Solution { public void reverseString(char[] s) { // if(s == null || s.length == 0) // return ""; int left = 0, right = s.length-1; while(left < right){ char temp = s[left]; s[left++] = s[right]; s[right--] = temp; } } }
151. Reverse Words in a String
1.反转全部字符数组
2.反转每个单词
3.清楚多余空格
class Solution { public String reverseWords(String s) { int n = s.length(); if (n == 0) return ""; char[] a = s.toCharArray(); //1. reverse string reverse(a, 0, n - 1); //2. reverse words reverseWord(a, n); //3. clean up spaces return cleanSpaces(a,n); } public String cleanSpaces(char[] a, int n){ int i = 0, j = 0; while(j < n){ while(j < n && a[j] == ‘ ‘) j++; while(j < n && a[j] != ‘ ‘) a[i++] = a[j++]; //keep non-spaces while(j < n && a[j] == ‘ ‘) j++; if(j < n) a[i++] = ‘ ‘; //keep only one space } return new String(a).substring(0, i); } void reverseWord(char[] a, int n){ int i = 0, j = 0; while(i < n){ while(i < j || i < n && a[i] == ‘ ‘) i++; //skip spaces while(j < i || j < n && a[j] != ‘ ‘) j++; //skip non-spaces reverse(a, i, j -1); } } private void reverse(char[] a, int i, int j){ while(i < j){ char temp = a[i]; a[i++] = a[j]; a[j--] = temp; } } }
该方法的作用是返回一个字符数组，该字符数组中存放了当前字符串中的所有字符。将字符串转换为字符数组。
2. 数组有length属性，String有length()方法。
优先级&& 大于 ||
标签：优先 ring clean 方法 tail pac color rds one
原文地址：https://www.cnblogs.com/Afei-1123/p/11783100.html
