1.描述：找出數組A的和最大的非空連續子數組，我們稱這樣的連續子數組為最大子數組。

2. 用分治策略來求解。

　　a. 假設我們要求A的子數組A[low, high]的最大子數組。根據分治策略，我們先將A[low,high] 平分

　　b. 那麽 A[low,highj]的子數組A[i,j]只有三種可能

　　　　a）完全位於A[low, mid]; 此時 low <= i <= j <= mid

　　　　b) 完全位於A[nid+1, high]中，此時 mid + 1 <= i <= j <= high

　　　　c) 跨越了中點mid， 此時 low <= i <= mid < j < high

3. 偽代碼

FIND-MAXIMUM-SUBARRAY(A, low, high)
    if high == low
        return (low, high. A[low])   //只有一個元素
    else 
        mid = (low + high)/2     //向下取整
        (left-low, left-high, left-sum) = FIND-MAXIMUM-SUBARRAY(A, low, mid)
        (right-low, right-high, right-sum) = FIND-MAXIMUM-SUBARRAY(A, mid + 1
 
, high)
        (cross-low, cross-high, cross-sum) = FIND-MAX-CROSSING-SUBARRAY(A, low, mid, high)
        if left-sum >= right-sum and left-sum >= cross-sum
            return (left-low, left-high, left-sum)
        else if right-sum >= left-sum and right-sum >= cross-sum
             return (right
 
-low, right-high, right-sum)
        return (cross-low, cross-high, cross-sum)


FIND-MAX-CROSSING-SUBARRAY(A, low, mid, high)
    left-sum = -∞
    sum = 0
    for i = mid downto low
        sum = sum + A[i]
        if sum > left-sum
            left-sum = sum
            max-left = i
    right-sum = -∞
    sum =0;
    for j = mid + 1 to high
        sum = sum + A[j]
        if sum > right-sum
            right-sum = sum
            max-right = j
    return (max-left, max-right, left-sum + right-sum)

4. 分析

　　我之前說過，所有的比較最後都是兩個數比較。把最大子數組通過分治策略最後都是一個元素，這時候就是直接返回這個數，交給上一層。

　　這時候數組有兩個數，子數組就到了2所說的比較三種情況，再一層層向上遞交結果

5. 代碼實現

java

public class MaxArray {

    private static class Result {
        int low;
        int high;
        int sum;

        public Result(int low, int high, int sum) {
            this.low = low;
            this.high = high;
            this.sum = sum;
        }
    }


    static Result findMaximumSubarray(int[] A, int low, int high) {
        if (low == high) {
            return new Result(low, high, A[low]);
        } else {
            int mid = (low + high)/2;
            Result leftResult = findMaximumSubarray(A, low, mid);
            Result rightResult = findMaximumSubarray(A, mid+1, high);
            Result crossResult = findMaxCrossingSubarray(A, low, mid, high);
            if (leftResult.sum >= rightResult.sum && leftResult.sum >= crossResult.sum)
                return leftResult;
            else if (rightResult.sum >= leftResult.sum && rightResult.sum >= crossResult.sum)
                return rightResult;
            else return crossResult;
        }

    }

    static Result findMaxCrossingSubarray(int[] A, int low, int mid, int high) {

        //向左試探
        int leftSum = Integer.MIN_VALUE;   //哨兵
        int maxLeft = mid;
        int sum = 0;
        for (int i = mid; i >= low; i--) {
            sum += A[i];
            if (sum > leftSum) {
                leftSum = sum;
                maxLeft = i;
            }
        }
        //向右試探
        int rightSum = Integer.MIN_VALUE;
        int maxRight = mid + 1;
        sum = 0;
        for (int j = mid + 1; j <= high; j++) {
            sum += A[j];
            if (sum > rightSum) {
                rightSum = sum;
                maxRight = j;
            }
        }
        //將兩邊的結果合起來
        return new Result(maxLeft, maxRight, leftSum + rightSum);
    }

    public static void main(String[] args) {
        int[] A = {-1, 5, 6, 9, 10, -9, -8, 100, -200};
        Result result = findMaximumSubarray(A, 0,  A.length-1);
        System.out.println(result.low + "," + result.high + " " + result.sum);
    }
}

python

def find_maximum_subarray(nums, low, high):
    if low == high:
        return {"low": low, "high": high, "sum": nums[low]}
    else:
        mid = int((low + high) / 2)
        left_result = find_maximum_subarray(nums, low, mid)
        right_result = find_maximum_subarray(nums, mid + 1, high)
        cross_result = find_max_crossing_subarray(nums, low, mid, high)
        if left_result["sum"] >= right_result["sum"] and left_result["sum"] >= cross_result["sum"]:
            return left_result
        else:
            if right_result["sum"] >= left_result["sum"] and right_result["sum"] >= cross_result["sum"]:
                return right_result
            else:
                return cross_result


def find_max_crossing_subarray(nums, low, mid, high):
    left_sum = -float(‘inf‘)
    total = 0
    max_left = mid
    for i in range(mid, low-1, -1):
        total += nums[i]
        if total > left_sum:
            left_sum = total
            max_left = i

    rigth_sum = -float(‘inf‘)
    total = 0
    max_right = mid + 1
    for j in range(mid+1, high+1):
        total += nums[j]
        if total > rigth_sum:
            rigth_sum = total
            max_right = j

    return {"low": max_left, "high": max_right, "sum": left_sum + rigth_sum}


if __name__ == "__main__":
    numss = [-1, 5, 6, 9, 10, -9, -8, 100, -200]
    result = find_maximum_subarray(numss, 0, len(numss)-1)
    print(result)

再分享個python用切片的方法

def find_maximum_subarray_slice(nums):
    max_sum = -float(‘inf‘)
    result = {}
    for i in range(len(nums)+1):
        for j in range(i, len(nums)+1):
            total = sum(nums[i:j])
            if total > max_sum:
                max_sum = total
                result["low"] = i
                result["high"] = j-1
    result["sum"] = max_sum
    return result

C語言

結構體沒有學好，晚點貼。。

【算法導論】最大子數組