最大子序和
1. 暴力法
工夫复杂度: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,顺次加上前面的数字,只存储最大值
-
依此类推
class Solution { public: int maxSubArray(vector<int> &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. 动静布局
工夫复杂度:O(N)
空间复杂度:O(1) - 一一加值比拟,存储最大值
-
一旦遇到加值后的后果 < 0,则只保留之前计算的最大值,从新开始下一个加值比拟
class Solution { public: int maxSubArray(vector<int> &nums) { int result = INT_MIN; int numsSize = int(nums.size()); //dp[i] 示意 nums 中以 nums[i] 结尾的最大子序和 vector<int> 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. 贪婪算法
工夫复杂度:O(N)
空间复杂度:O(1)
与动静布局基本一致class Solution { public: int maxSubArray(vector<int> &nums) { // 相似寻找最大最小值的题目,初始值肯定要定义成实践上的最小最大值 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,从新开始找子序串 if (sum < 0) {sum = 0;} } return result; } };
4. 分治法
工夫复杂度:O(nlog(n))
空间复杂度:O(log(n))
class Solution
{
public:
int maxSubArray(vector<int> &nums)
{
// 相似寻找最大最小值的题目,初始值肯定要定义成实践上的最小最大值
int result = INT_MIN;
int numsSize = int(nums.size());
result = maxSubArrayHelper(nums, 0, numsSize - 1);
return result;
}
int maxSubArrayHelper(vector<int> &nums, int left, int right)
{if (left == right)
{return nums[left];
}
int mid = (left + right) / 2;
int leftSum = maxSubArrayHelper(nums, left, mid);
// 留神这里应是 mid + 1,否则 left + 1 = right 时,会有限循环
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<int> &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;
// 留神这里 i = mid + 1,防止反复用到 nums[i]
for (int i = mid + 1; i <= right; i++)
{sum += nums[i];
rightSum = max(rightSum, sum);
}
return (leftSum + rightSum);
}
};