53. Maximum Subarray
Given an integer array nums
, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
方法一
public int maxSubArray(int[] nums) { int max=nums[0]; int total=0; for(int i=0;i<nums.length;i++) { if(total<0) { total=nums[i]; } else { total+=nums[i]; } if(total>max) { max=total; } } return max; }
方法2 动态规划
public int maxSubArray(int[] nums) { int[] dp = new int[nums.length]; dp[0]=nums[0]; int max=dp[0]; for(int i=1;i<nums.length;i++) { dp[i] =nums[i]+(dp[i-1]>0?dp[i-1]:0); max=Math.max(max,dp[i]); } return max; }
58. Length of Last Word
Given a string s consists of upper/lower-case alphabets and empty space characters ' '
, return the length of last word in the string.
If the last word does not exist, return 0.
Note: A word is defined as a character sequence consists of non-space characters only.
Example:
Input: "Hello World" Output: 5
方法一
public int lengthOfLastWord(String s) { if(s.trim().length()==0) { return 0; } int len=0; String[] ss=s.split(" "); return ss[ss.length-1].length(); }
改进
public int lengthOfLastWord(String s) { return s.trim().length()-s.trim().lastIndexOf(" ")-1; }
方法 2
public int lengthOfLastWord(String s) { if(s.trim().length()==0) { return 0; } int count=0; int i=s.length()-1; while(i>=0&&(s.charAt(i--)==' ')); i++; while(i>=0&&(s.charAt(i--)!=' ')) { count++; } return count; }