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 == 1) {
            return nums[0];
        }
        int *data = nums.data();
        return helper(data, size);
    }
    int helper(int *data, int n) {
        if ( n == 1) {
            return data[0];
        }
        int mid = n >> 1;
        int ans = max(helper(data, mid), helper(data + mid, n - mid));
        int now = data[mid - 1], may = now;
        for (int i = mid - 2; i >= 0; 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 == 1) {
            return nums[0];
        }
        vector<int> dp(size);
        dp[0] = nums[0];
        int ans = dp[0];
        for (int i=1; i<size; i++) {
            dp[i] = max(dp[i - 1] + 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 == 1) {
            return nums[0];
        }
        int endHere = nums[0];
        int ans = nums[0];
        for (int i=1; 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 == 1) {
            return nums[0];
        }
        int sum = nums[0];
        int minSum = min(0, sum);
        int ans = nums[0];
        for (int i = 1; i < size; ++i) {
            sum += nums[i];
            ans = max(ans, sum - minSum);
            minSum = min(minSum, sum);
        }
        return ans;
    }
};