【力扣】 - 53. 最大子序和力扣

最大子序和 1. 暴力法

时间复杂度：O(N^2)
空间复杂度：O(1)

设置两层for循环
存储第一个数字的值，依次加上后面的数字，只存储最大值
依此类推

class Solution { public: int maxSubArray(vector &nums) { int max = INT_MIN; int numsSize = int(nums.size()); for (int i = 0; i < numsSize; i++) { int sum = 0; for (int j = i; j < numsSize; j++) { sum += nums[j]; if (sum > max) { max = sum; } } } return max; } };

2. 动态规划
时间复杂度：O(N)
空间复杂度：O(1)
逐个加值比较，存储最大值
【【力扣】 - 53. 最大子序和】一旦遇到加值后的结果 < 0，则只保留之前计算的最大值，重新开始下一个加值比较

class Solution { public: int maxSubArray(vector &nums) { int result = INT_MIN; int numsSize = int(nums.size()); //dp[i]表示nums中以nums[i]结尾的最大子序和 vector dp(numsSize); dp[0] = nums[0]; result = dp[0]; for (int i = 1; i < numsSize; i++) { dp[i] = max(dp[i - 1] + nums[i], nums[i]); result = max(result, dp[i]); } return result; } };

3.贪心算法
时间复杂度：O(N)
空间复杂度：O(1)
与动态规划基本一致
class Solution { public: int maxSubArray(vector &nums) { //类似寻找最大最小值的题目，初始值一定要定义成理论上的最小最大值 int result = INT_MIN; int numsSize = int(nums.size()); int sum = 0; for (int i = 0; i < numsSize; i++) { sum += nums[i]; result = max(result, sum); //如果sum < 0，重新开始找子序串 if (sum < 0) { sum = 0; } }return result; } };

4. 分治法
时间复杂度：O(nlog(n))
空间复杂度：O(log(n))

class Solution { public: int maxSubArray(vector &nums) { //类似寻找最大最小值的题目，初始值一定要定义成理论上的最小最大值 int result = INT_MIN; int numsSize = int(nums.size()); result = maxSubArrayHelper(nums, 0, numsSize - 1); return result; }int maxSubArrayHelper(vector &nums, int left, int right) { if (left == right) { return nums[left]; } int mid = (left + right) / 2; int leftSum = maxSubArrayHelper(nums, left, mid); //注意这里应是mid + 1，否则left + 1 = right时，会无限循环 int rightSum = maxSubArrayHelper(nums, mid + 1, right); int midSum = findMaxCrossingSubarray(nums, left, mid, right); int result = max(leftSum, rightSum); result = max(result, midSum); return result; }int findMaxCrossingSubarray(vector &nums, int left, int mid, int right) { int leftSum = INT_MIN; int sum = 0; for (int i = mid; i >= left; i--) { sum += nums[i]; leftSum = max(leftSum, sum); }int rightSum = INT_MIN; sum = 0; //注意这里i = mid + 1，避免重复用到nums[i] for (int i = mid + 1; i <= right; i++) { sum += nums[i]; rightSum = max(rightSum, sum); } return (leftSum + rightSum); } };