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;
}