LintCode: Maximum Subarray

1. 暴力枚举

2. “聪明”枚举

3. 分治法

分：两个基本等长的子数组，分别求解T(n/2)

合：跨中心点的最大子数组合（枚举）O(n)

时间复杂度：O(n*logn)

 1 class Solution {
 2 public:    
 3     /**
 4      * @param nums: A list of integers
 5      * @return: A integer indicate the sum of max subarray
 6      */
 7     int maxSubArray(vector<int> nums) {
 8         // write your code here
 9         int size = nums.size();
10         if (size == 1) {
11             return nums[0];
12         }
13         int *data = nums.data();
14         return helper(data, size);
15     }
16     int helper(int *data, int n) {
17         if ( n == 1) {
18             return data[0];
19         }
20         int mid = n >> 1;
21         int ans = max(helper(data, mid), helper(data + mid, n - mid));
22         int now = data[mid - 1], may = now;
23         for (int i = mid - 2; i >= 0; i--) {
24             may = max(may, now += data[i]);
25         }
26         now = may;
27         for (int i = mid; i < n; i++) {
28             may = max(may, now += data[i]);
29         }
30         return max(ans, may);
31     }
32 };

 

4. dp（不枚举子数组，枚举方案）

dp[i]表示以a[i]结尾的最大子数组的和

dp[i] = max(dp[i-1]+a[i], a[i])

　　包含a[i-1]:dp[i-1]+a[i]

　　不包含a[i-1]:a[i]

初值：dp[0] = a[0]

答案：最大的dp[0...n-1]

时间：O(n)

空间：O(n)

空间优化：dp[i]要存吗？

　　endHere = max(endHere+a[i], a[i])

　　answer = max(endHere, answer)

优化后的空间：O(1)

 1 class Solution {
 2 public:    
 3     /**
 4      * @param nums: A list of integers
 5      * @return: A integer indicate the sum of max subarray
 6      */
 7     int maxSubArray(vector<int> nums) {
 8         // write your code here
 9         int size = nums.size();
10         if (size == 1) {
11             return nums[0];
12         }
13         vector<int> dp(size);
14         dp[0] = nums[0];
15         int ans = dp[0];
16         for (int i=1; i<size; i++) {
17             dp[i] = max(dp[i - 1] + nums[i], nums[i]);
18             ans = max(ans, dp[i]);
19         }
20         return ans;
21     }
22 };

空间优化

 1 class Solution {
 2 public:    
 3     /**
 4      * @param nums: A list of integers
 5      * @return: A integer indicate the sum of max subarray
 6      */
 7     int maxSubArray(vector<int> nums) {
 8         // write your code here
 9         int size = nums.size();
10         if (size == 1) {
11             return nums[0];
12         }
13         int endHere = nums[0];
14         int ans = nums[0];
15         for (int i=1; i<size; i++) {
16             endHere = max(endHere + nums[i], nums[i]);
17             ans = max(ans, endHere);
18         }
19         return ans;
20     }
21 };

 

5. 另外一种线性枚举

定义：sum[i] = a[0] + a[1] + a[2] + ... + a[i]  i>=0

　　   sum[-1] = 0

则对0<=i<=j：

　　a[i] + a[i+1] + ... + a[j] = sum[j] - sum[i-1]

我们就是要求这样一个最大值：

　　对j我们可以求得当前的sum[j],取的i-1一定是之前最小的sum值，用一个变量记录sum的最小值

　　时间：O(n)

　　空间：O(1)

 1 class Solution {
 2 public:    
 3     /**
 4      * @param nums: A list of integers
 5      * @return: A integer indicate the sum of max subarray
 6      */
 7     int maxSubArray(vector<int> nums) {
 8         // write your code here
 9         int size = nums.size();
10         if (size == 1) {
11             return nums[0];
12         }
13         int sum = nums[0];
14         int minSum = min(0, sum);
15         int ans = nums[0];
16         for (int i = 1; i < size; ++i) {
17             sum += nums[i];
18             ans = max(ans, sum - minSum);
19             minSum = min(minSum, sum);
20         }
21         return ans;
22     }
23 };

 

posted @ 2015-11-30 10:22  ZH奶酪  阅读(250)  评论(0)  编辑  收藏  举报