Leetcode剑指Offer刷题指南：

Leetcode剑指Offer刷题-学习计划目录_DEGv587的博客-CSDN博客

剑指 Offer 21. 调整数组顺序使奇数位于偶数前面

解法：双指针交换（易理解版）

class Solution {public int[] exchange(int[] nums) {int start = 0, end = nums.length - 1;while (start < end - 1) {//start处为奇数时往后走if (nums[start] % 2 != 0) {start++;}//end处为偶数时往前走if (nums[end] % 2 == 0) {end--;}//start处为偶数，end处为奇数时，进行交换if (nums[start] % 2 == 0 && nums[end] % 2 != 0) {swap(nums, start, end);}}return nums;}private void swap(int[] nums, int start, int end) {int tmp = nums[start];nums[start] = nums[end];nums[end] = tmp;}
}

 优化版：

class Solution {public int[] exchange(int[] nums) {int start = 0, end = nums.length - 1;while (start <= end) {if (nums[start] % 2 != 0) {start++;} else if (nums[end] % 2 == 0) {end--;} else {swap(nums, start, end);}}return nums;}private void swap(int[] nums, int start, int end) {int tmp = nums[start];nums[start] = nums[end];nums[end] = tmp;}
}

简化版：

class Solution {public int[] exchange(int[] nums) {int start = 0, end = nums.length - 1;while (start <= end) {if (nums[start] % 2 == 1) {start++;} else if (nums[end] % 2 == 0) {end--;} else {int tmp = nums[start];nums[start] = nums[end];nums[end] = tmp;}}return nums;}
}

剑指 Offer 57. 和为s的两个数字

解法一：HashSet

class Solution {public int[] twoSum(int[] nums, int target) {HashSet<Integer> set = new HashSet<>();for (int x : nums) {if (!set.contains(target - x)) {set.add(x);} else {return new int[]{x, target - x};}}return null;}
}

解法二：二分法

class Solution {public int[] twoSum(int[] nums, int target) {for (int i = 0; i < nums.length; ++i) {int x = target - nums[i];int left = i, right = nums.length - 1;while (left <= right) {int mid = (left + right) >> 1;if (x < nums[mid]) {right = mid - 1;} else if (x > nums[mid]) {left = mid + 1;} else {return new int[]{nums[i], nums[mid]};}}}return null;}
}

解法三：双指针

class Solution {public int[] twoSum(int[] nums, int target) {int left = 0, right = nums.length - 1;while (left < right) {int sum = nums[left] + nums[right];if (sum < target) {left++;} else if (sum > target) {right--;} else {return new int[]{nums[left], nums[right]};}}return null;}
}

剑指 Offer 58 - I. 翻转单词顺序

解法一：分割 + 倒序

class Solution {public String reverseWords(String s) {String[] strs = s.trim().split(" ");StringBuilder sb = new StringBuilder();for (int i = strs.length - 1; i >= 0; --i) {if (strs[i].equals("")) {continue;}sb.append(strs[i].trim());if (i == 0) {break;} else {sb.append(" ");}}return sb.toString();}
}

解法二：双指针

class Solution {public String reverseWords(String s) {//去掉前后空格s = s.trim();StringBuilder sb = new StringBuilder();int left = s.length() - 1, right = left;while (left >= 0) {//找到第一个空格位置while (left >= 0 && s.charAt(left) != ' ') {left--;}//添加找到的第一个空格后的第一个单词sb.append(s.substring(left + 1, right + 1)).append(" ");//跳过中间空格，找到空格前的位置while (left >= 0 && s.charAt(left) == ' ') {left--;}right = left;}return sb.toString().trim();}
}