leetcode 53. Maximum Subarray最大连续子序列和

最大连续子序列和

leetcode 53. Maximum Subarray

动态规划

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        vector<int> d(nums.size(),0);
        d[0]=nums[0];
        for(size_t i=1;i1]+nums[i]);
        int maxsum=d[0];
        for(size_t i=1;ireturn maxsum;
    }
};

分治

原始连接

mx (largest sum of this subarray), 
lmx(largest sum starting from the left most element), 
rmx(largest sum ending with the right most element), 
sum(the sum of the total subarray). 

class Solution {
public:
    void maxSubArray(vector<int>& nums, int l, int r, int& mx, int& lmx, int& rmx, int& sum) {
        if (l == r) {
            mx = lmx = rmx = sum = nums[l];
        }
        else {
            int m = (l + r) / 2;
            int mx1, lmx1, rmx1, sum1;
            int mx2, lmx2, rmx2, sum2;
            maxSubArray(nums, l, m, mx1, lmx1, rmx1, sum1);
            maxSubArray(nums, m + 1, r, mx2, lmx2, rmx2, sum2);
            mx = max(max(mx1, mx2), rmx1 + lmx2);
            lmx = max(lmx1, sum1 + lmx2);
            rmx = max(rmx2, sum2 + rmx1);
            sum = sum1 + sum2;
        }
    }
    int maxSubArray(vector<int>& nums) {
        if (nums.size() == 0) {
            return 0;
        }
        int mx, lmx, rmx, sum;
        maxSubArray(nums, 0, nums.size() - 1, mx, lmx, rmx, sum);
        return mx;
    }
};