leetcode 53,58

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;
    }