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)

 class Solution {
 public:
     /**
      * @param nums: A list of integers
      * @return: A integer indicate the sum of max subarray
      */
     int maxSubArray(vector<int> nums) {
         // write your code here
         int size = nums.size();
         if (size == ) {
             return nums[];
         }
         int *data = nums.data();
         return helper(data, size);
     }
     int helper(int *data, int n) {
         if ( n == ) {
             return data[];
         }
         int mid = n >> ;
         int ans = max(helper(data, mid), helper(data + mid, n - mid));
         int now = data[mid - ], may = now;
         for (int i = mid - ; i >= ; i--) {
             may = max(may, now += data[i]);
         }
         now = may;
         for (int i = mid; i < n; i++) {
             may = max(may, now += data[i]);
         }
         return max(ans, may);
     }
 };

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)

 class Solution {
 public:
     /**
      * @param nums: A list of integers
      * @return: A integer indicate the sum of max subarray
      */
     int maxSubArray(vector<int> nums) {
         // write your code here
         int size = nums.size();
         if (size == ) {
             return nums[];
         }
         vector<int> dp(size);
         dp[] = nums[];
         int ans = dp[];
         for (int i=; i<size; i++) {
             dp[i] = max(dp[i - ] + nums[i], nums[i]);
             ans = max(ans, dp[i]);
         }
         return ans;
     }
 };

空间优化

 class Solution {
 public:
     /**
      * @param nums: A list of integers
      * @return: A integer indicate the sum of max subarray
      */
     int maxSubArray(vector<int> nums) {
         // write your code here
         int size = nums.size();
         if (size == ) {
             return nums[];
         }
         int endHere = nums[];
         int ans = nums[];
         for (int i=; i<size; i++) {
             endHere = max(endHere + nums[i], nums[i]);
             ans = max(ans, endHere);
         }
         return ans;
     }
 };

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)

 class Solution {
 public:
     /**
      * @param nums: A list of integers
      * @return: A integer indicate the sum of max subarray
      */
     int maxSubArray(vector<int> nums) {
         // write your code here
         int size = nums.size();
         if (size == ) {
             return nums[];
         }
         int sum = nums[];
         int minSum = min(, sum);
         int ans = nums[];
         for (int i = ; i < size; ++i) {
             sum += nums[i];
             ans = max(ans, sum - minSum);
             minSum = min(minSum, sum);
         }
         return ans;
     }
 };