EP42-Dynamic Programming2

12/04/2017更新

Practice makes perfect. 古人诚不欺我也。今天无意间回顾了一下这道题，发现当年的思路太混乱了，State transition equation写的也不对。而且代码写得太「野路子」了。
详见：http://www.jianshu.com/p/46ea9e851b41

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Yesterday we talked about some basic theories about dynamic programming, honestly I'm still not very confident about that, just got a implicit concept in my mind. Here are some other classic problems about DP, maybe we can grab something from them.

Maximum SubArray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

Ok this is another typical Dynamic Programming issue. Like I mentioned yesterday, the most difficult/important part of using DP to solve a problem, is to recognize that this problem can be solved by the idea of DP. So how can we know that it can be solve by DP?
You can see the first number of the array is -2, the second number is 1,then the sum of the subarray consisting of the first 2 numbers is -1 + 2 = -1 , however -1 cannot contribute to the maxSubarray because it will make the sum smaller, so we just discard -1, and just start from scratch: make sum = 0.
The state transition equation is:

Max[i+1]
= Max[i] + A[i+1], if (Max[i] + A[i+1] >0)
= 0 ,if (Max[i] + A[i+1] >0) (it shall not be Max[i])

public int maxSubArray(int[] nums) {
    int sum = nums[0], maxSum = nums[0];
    for (int i = 0; i < nums.length; i++) {
        if (sum < 0) {
            sum = 0;
        }
        sum += nums[i];
        //find a new maxSum which saves the current maximum sum
        maxSum = Math.max(maxSum, sum);
    }
    return maxSum;
}

http://www.tuicool.com/articles/IbiMjaI
http://www.tuicool.com/articles/UzmU7jb