[leetcode]53. Maximum Subarray

被44题虐了，先跳个easy缓一缓。easy题就是easy题啊

Solution 1:自己的思路,但是单独处理了全负数的情况

class Solution {
    public int maxSubArray(int[] nums) {
        int i=0;
        int sum=0;
        int max=0;
        int flag=0;
        
        //处理全为负数的情况
        for(int j=0;j<nums.length;j++){
          if(nums[j]>=0){
              flag=1;
              break;
          }
        }
        if(flag==0) {
            Arrays.sort(nums);
            return nums[nums.length-1];
        }
        
        while(i<nums.length){
            if(sum+nums[i]<0){//只要加完小于0，那么sum归0，从头开始
                sum=0;
            }
            else{//否则，无论正数，负数都加上去再说
                sum+=nums[i];
                
                if(sum>=max){  
                    max=sum;
                }
            }
            
            i++;
        }
        
        return max;
    }
}

Solution 2:

从solution2-4，其实都是一个本质的思想

class Solution {
    public int maxSubArray(int[] nums) {
        int i=1;
        int sum=nums[0];
        int max=nums[0];
       
        
        while(i<nums.length){
            if(sum+nums[i]<nums[i]){
                sum=nums[i];
            }
            else{
                sum+=nums[i];
            }
           
            max=Math.max(sum,max);
            i++;
        }
        
        return max;
    }
}

Solution 3: 2的精简版

  public int maxSubArray(int[] nums) {
        if(nums.length == 0) {
            return 0;
        }
        int curMax = nums[0];
        int allMax = nums[0];
        for(int i = 1; i < nums.length; i++) {
            curMax = Math.max(nums[i], curMax + nums[i]);
            allMax = Math.max(allMax, curMax);
        }
        return allMax;
    }

Solution 4: dp

public int maxSubArray(int[] A) {
        int n = A.length;
        int[] dp = new int[n];//dp[i] means the maximum subarray ending with A[i];
        dp[0] = A[0];
        int max = dp[0];
        
        for(int i = 1; i < n; i++){
            dp[i] = A[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);
            max = Math.max(max, dp[i]);
        }
        
        return max;
}