53. Maximum Subarray

最近迷恋DP，我看还是将这个主题先解决一些。

Problem

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.

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

求最大连续子序列问题

思路：

dp[i] = max{dp[i-1]+a[i], a[i]};

• 包含a[i-1]，dp[i] = dp[i-1] + a[i];
• 不包含a[i]，dp[i] = a[i];

初值：dp[0] = a[0];

答案：遍历dp，找出最大。

时间复杂度O{n}，空间复杂度O{n}

空间复杂度为N

class Solution {public:    int maxSubArray(vector<int>& nums) {        int n = nums.size();        vector<int> dp(n);        int answer;        dp[0] = nums[0];        answer = dp[0];        for (int i = 1; i < n; ++i) {            dp[i] = max(dp[i-1]+nums[i], nums[i]);            if (answer < dp[i]) {                answer = dp[i];            }        }        return answer;    }};

空间复杂度为1

在求answer的过程中，不需要保存所有dp

class Solution {public:    int maxSubArray(vector<int>& nums) {        int n = nums.size();        int dp = nums[0], answer = nums[0];        for (int i = 1; i < n; ++i) {            dp = max(dp+nums[i], nums[i]);            //answer = max(dp, answer);            answer = (answer < dp ? dp : answer);         }        return answer;    }};