[leetcode] 53. Maximum Subarray
2016-08-12 11:59
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Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array
the contiguous subarray
解法一:
O(n)的解法。用一个global表示全局最大,一个local表示局部最大。那么局部最大的表达式是local = max(local+nums[i], nums[i]),也就是如果之前的局部最大为负数,不如不要,直接只用nums[i]。然后全局最优就是之前global和当前局部最优的max。
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int global = nums[0], local = nums[0];
for(int i=1; i<nums.size(); i++){
local = max(local+nums[i], nums[i]);
global = max(global, local);
}
return global;
}
};
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.
解法一:
O(n)的解法。用一个global表示全局最大,一个local表示局部最大。那么局部最大的表达式是local = max(local+nums[i], nums[i]),也就是如果之前的局部最大为负数,不如不要,直接只用nums[i]。然后全局最优就是之前global和当前局部最优的max。
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int global = nums[0], local = nums[0];
for(int i=1; i<nums.size(); i++){
local = max(local+nums[i], nums[i]);
global = max(global, local);
}
return global;
}
};
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