【LeetCode】053. Maximum Subarray

题目：

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.

题解：

　　遍历所有组合，更新最大和。

Solution 1 (TLE)

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

　　可以看做动态规划的简略版，见Solution 5

Solution 2 ()

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

　　贪心算法：The idea is to find the largest difference between the sums when you summing up the array from left to right. The largest difference corresponds to the sub-array with largest sum

Solution 3 ()

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int sum = 0, minsum = 0, result = nums[0], n = nums.size();
        for(int i = 0; i < n; i++) {
            sum += nums[i];
            if(sum - minsum > res) result = sum - minsum;
            if(sum < minsum) minsum = sum;
        }
        return result;
    }
};

　　分治法：

Solution 4 ()

　　DP算法：维护一个一维数组。

Solution 5 ()

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp(n,0);//dp[i] means the maximum subarray ending with nums[i];
        dp[0] = nums[0];
        int max = dp[0];
        
        for(int i = 1; i < n; i++){
            dp[i] = nums[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);
            max = Math.max(max, dp[i]);
        }        
        return max;
    }
};