排序算法

快速排序
归并排序
冒泡排序

查找算法

字符串

LeetCode 125. 验证回文串

 public static boolean isPalindrome(String s) {
    int right = s.length() - 1;
    int left = 0;
    while (left < right) {
      char leftChar = s.charAt(left);
      while (!Character.isLetterOrDigit(leftChar) && left < right) {
        leftChar = s.charAt(++left);
      }
      char rightChar = s.charAt(right);
      while (!Character.isLetterOrDigit(rightChar) && left < right) {
        rightChar = s.charAt(--right);
      }
      if (Character.toLowerCase(leftChar) != Character.toLowerCase(rightChar)) {
        return false;
      }
      left++;
      right--;
    }
    return true;
  }

数组相关

LeetCode 1. 两数之和

  public static int[] twoSum(int[] nums, int target) {
    for (int i = 0; i < nums.length; i++) {
      int result = target - nums[i];
      for (int j = i + 1; j < nums.length; j++) {
        if (nums[j] == result) {
          return new int[] { i, j };
        }
      }
    }
    return null;
  }

 public static int[] twoSum1(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
      if (map.containsKey(target - nums[i])) {
        return new int[] { map.get(target - nums[i]), i };
      }
      map.put(nums[i], i);
    }
    return null;
  }

303. 区域和检索 - 数组不可变

// 解法1
class NumArray {
    private final int[] nums;
    public NumArray(int[] nums) {
       this.nums=nums;
    }
    
    public int sumRange(int left, int right) {
        if(nums==null){
            return 0;
        }
        int sum = 0;
        for(;left<=right;left++){
            sum += nums[left];
        }
        return sum;
    }
}

// 解法2
public class NumArray {
    private final int[] sums;

    public NumArray(int[] nums) {
        int n = nums.length;
        sums = new int[n + 1];
        for (int i = 0; i < n; i++) {
            sums[i + 1] = sums[i] + sums[i + 1];
        }
    }

    public int sumRange(int i, int j) {
        return sums[j + 1] - sums[i];
    }
}