leetcode01-最大子序和

1.題目描述：

給定一個整數數組nums,找到一個具有最大和的連續子數組（子數組最少包含一個元素），返回其最大和；

2.來源：

https://leetcode-cn.com/problems/maximum-subarray/

3.編吧：

動態規劃

核心思想：若當前元素之前的和小於0(presum<0)，則丟棄當前元素之前的數列；

步驟：初始值:maxsum = nums[0] = -2;presum = 0

（1）num = -2;

presum = max(presum+num,num) = max(0-2,-2) = -2;

maxsum = max(maxsum,presum) = max(-2,-2) = -2;

（此時子數組從第一個元素-2開始）

（2）num = 1;

presum = max(presum+num,num) = max(-2+1,1) = 1;

maxsum = max(maxsum,presum) = max(-2,1) = 1;

(由於上一步驟計算出的presum<0;所以丟棄前面的數列，子數組從第二個元素1開始)

（3）num = -3,

presum = max(presum+num,num) = max(1-3,-3) = -2;

maxsum = max(maxsum,presum) = max(1,-2) = 1;

（4）num = 4,

presum = max(presum+num,num) = max(-2+4，4) = 4;

maxsum = max(maxsum,presum) = max(1,2) = 4;

(由於上一步驟計算出的presum<0;所以丟棄前面的數列，子數組從第四個元素4開始)

（5）num = -1;

presum = max(presum+num,num) = max(4-1，-1) = 3;

maxsum = max(maxsum,presum) = max(4，3) = 4;

（6）num = 2;

presum = max(presum+num,num) = max(3+2，2) = 5;

maxsum = max(maxsum,presum) = max(4,5) = 5;

（7）num = 1;

presum = max(presum+num,num) = max(5+1，1) = 6;

maxsum = max(maxsum,presum) = max(5,6) = 6;

（8）num = -5;

presum = max(presum+num,num) = max(6-5,-5) = 1;

maxsum = max(maxsum,presum) = max(6,1) = 6;

（9）num = 4;

presum = max(presum+num,num) = max(1+4,4) = 5;

maxsum = max(maxsum,presum) = max(6,5) = 6;

總結：從第（7）步驟開始max的值始終爲6；並且子數組從第四個元素4開始，到第七個元素1結束，即數組[4,-1,2,1]爲nums數組中具有最大和的連續子數組；

4.Code Show

class Solution {
    public int maxSubArray(int[] nums) {
        int max = nums[0];
        int before = 0;
        for(int num:nums){
            before = Math.max(before+num,num);
            max = Math.max(max,before);
        }
        return max;
    }
}

 

5.複雜度分析