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.

我的思路：沒有一丟丟思路就是了

網上的思路：this problem was discussed by Jon Bentley (Sep. 1984 Vol. 27 No. 9 Communications of the ACM P885)

the paragraph below was copied from his paper (with a little modifications)

algorithm that operates on arrays: it starts at the left end (element A[1]) and scans through to the right end (element A[n]), keeping track of the maximum sum subvector seen so far. The maximum is initially A[0]. Suppose we've solved the problem for A[1 .. i - 1]; how can we extend that to A[1 .. i]? The maximum

sum in the first I elements is either the maximum sum in the first i - 1 elements (which we'll call MaxSoFar), or it is that of a subvector that ends in position i (which we'll call MaxEndingHere).

MaxEndingHere is either A[i] plus the previous MaxEndingHere, or just A[i], whichever is larger.

網上的程式碼：

public static int maxSubArray(int[] A) { int maxSoFar=A[0], maxEndingHere=A[0]; for (int i=1;i<A.length;++i){ maxEndingHere= Math.max(maxEndingHere+A[i],A[i]); maxSoFar=Math.max(maxSoFar, maxEndingHere); } return maxSoFar; }

自己看了後寫成了一模一樣的了 囧