53 Maximum Subarray

Author: int32bit

README.md

@@ -34,6 +34,7 @@
 + [47 Permutations II(全排列)](algorithms/Permutations2)
 + [49 Anagrams(字符串，回文构词)](algorithms/Anagrams)
 + [50 Pow(x, n)(二分，浮点数比较，INT\_MIN绝对值)](algorithms/Pow)
++ [53 Maximum Subarray(DP, 最大连续之和)](algorithms/MaximumSubarray)
 + [58 Length of Last Word](algorithms/LengthofLastWord)
 + [60 Permutation Sequence(康托展开)](algorithms/PermutationSequence)
 + [64 Minimum Path Sum(DP,空间压缩)](algorithms/MinimumPathSum)
@@ -89,7 +90,7 @@
 + [148 Sort List(归并排序)](algorithms/SortList)
 + [150 Evaluate Reverse Polish Notation(栈的应用)](algorithms/EvaluateReversePolishNotation)
 + [151 Reverse Words in a String](algorithms/ReverseWordsinaString)
-+ [152 Maximum Product Subarray(DP)](algorithms/MaximumProductSubarray)
++ [152 Maximum Product Subarray(DP,最大连续之积)](algorithms/MaximumProductSubarray)
 + [153 Find Minimum in Rotated Sorted Array(二分)](algorithms/FindMinimuminRotatedSortedArray)
 + [154 Find Minimum in Rotated Sorted Array II(二分)](algorithms/FindMinimuminRotatedSortedArray2)
 + [155 Min Stack(数据结构设计、栈的使用)](algorithms/MinStack)

algorithms/MaximumProductSubarray/README.md

@@ -40,3 +40,7 @@ int maxProduct(vector<int>& nums) {
     return max;
 }
 ```
+
+## 扩展
+
+[Maximum Subarray](../MaximumSubarray): 最大连续之和

algorithms/MaximumSubarray/README.md

@@ -0,0 +1,35 @@
+## 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.
+
+click to show more practice.
+More practice:
+
+If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
+
+## Solution
+
+最大连续之和经典题
+
+```c
+int maxSubArray(int *a, int n)
+{
+	if (a == NULL || n <= 0)
+		return 0;
+	int max = a[0], sum = a[0];
+	for (int i = 1; i < n; ++i) {
+		if (sum < 0)
+			sum = a[i];
+		else
+			sum += a[i];
+		max = MAX(max, sum);
+	}
+	return max;
+}
+```
+## 扩展
+
+[Maximum Product Subarray](../MaximumProductSubarray): 最大连续之积

algorithms/MaximumSubarray/in.txt

@@ -0,0 +1,6 @@
+9
+-2 1 -3 4 -1 2 1 -5 4
+5
+-1 -2 -3 -4 -5
+5
+1 -8 2 3 4

algorithms/MaximumSubarray/solve.c

@@ -0,0 +1,38 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+static inline int MAX(int a, int b)
+{
+	return a > b ? a : b;
+}
+int maxSubArray(int *a, int n)
+{
+	if (a == NULL || n <= 0)
+		return 0;
+	int max = a[0], sum = a[0];
+	for (int i = 1; i < n; ++i) {
+		if (sum < 0)
+			sum = a[i];
+		else
+			sum += a[i];
+		max = MAX(max, sum);
+	}
+	return max;
+}
+void print(int *a, int n)
+{
+	for (int i = 0; i < n; ++i)
+		printf("%d ", a[i]);
+	printf("\n");
+}
+int main(int argc, char **argv)
+{
+	int a[20], n;
+	memset(a, 0, sizeof(a));
+	while (scanf("%d", &n) != EOF) {
+		for (int i = 0; i < n; ++i)
+			scanf("%d", &a[i]);
+		printf("%d\n", maxSubArray(a, n));
+	}
+	return 0;
+}