剑指Offer之Java算法习题精讲数组与列表的查找及字符串转换

题目一

解法

class Solution {
  public String toLowerCase(String s) {
      StringBuilder sb = new StringBuilder();
      for(int i = 0;i<s.length();i++){
          char ch = s.charAt(i);
          if('A'<=ch&&ch<='Z'){
              ch = (char)(ch+32);
          }
          sb.append(ch);
      }
      return sb.toString();
  }
}

 

题目二

解法

class Solution {
  public int pivotIndex(int[] nums) {
      int sum = 0;
      for(int i = 0;i<nums.length;i++){
          sum+=nums[i];
      }
      int left = 0;
      int right = sum;
      for(int i = 0;i<nums.length;i++){
          right = right-nums[i];
          if(i == 0){
              left = 0;
              if(right==left) return 0;
          }else{
              left+=nums[i-1];
              if(right==left){
                  return i;
              }
          }
      }
      return -1;
  }
}

 

 题目三

 解法

class Solution {
  public List<Integer> selfDividingNumbers(int left, int right) {
      ArrayList<Integer> list = new ArrayList<Integer>();
      for(int i = left;i<=right;i++){
          if(selfDividing) list.add(i);
      }
      return list;
  }
  public boolean selfDividing(int n) {
      for (char c: String.valueOf(n).toCharArray()) {
          if (c == '0' || (n % (c - '0') > 0))
              return false;
      }
      return true;
  }   
}

 

题目四

 解法

class Solution {
  public char nextGreatestLetter(char[] letters, char target) {
      int left = 0;
      int right = letters.length;
      while(left<right){
          int mid = left+(right-left)/2;
          if(letters[mid]<=target){
              left = mid+1;
          }else{
              right = mid;
          }
      }
      return left==letters.length?letters[0]:letters[left];
  }
}

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原文链接：https://blog.csdn.net/wai_58934/article/details/123291730

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