Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
A subarray is a contiguous part of an array. Example 1
Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Example 2
Input: nums = [1]
Output: 1
Example 3
Input: nums = [5,4,-1,7,8]
Output: 23
Constraints
- 1 <= nums.length <= 10e5
- -10e4 <= nums[i] <= 10e4 Follow up: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
Solution
第一次尝试：WA
负数用例过不了，因为小于零直接就退出了，输出的是 maxSum 的初始值
int maxSubArray(int* nums, int numsSize){
int i, thisSum, maxSum;
thisSum = 0;
maxSum = 0x80000000;
for (i = 0; i < numsSize; i++) {
thisSum += nums[i];
if (thisSum < 0)
thisSum = 0;
else if (thisSum > maxSum)
maxSum = thisSum;
}
return maxSum;
}
第二次尝试：WA
这次先判断 maxSum 再判断大于零，先负数再正数错误，没有更新
int maxSubArray(int* nums, int numsSize){
int i, thisSum, maxSum;
thisSum = 0;
maxSum = 0x80000000;
for (i = 0; i < numsSize; i++) {
thisSum += nums[i];
if (thisSum > maxSum)
maxSum = thisSum;
else if (thisSum < 0)
thisSum = 0;
}
return maxSum;
}
第三次尝试：AC
解决方案：把分支去掉，两个 if 就行
int maxSubArray(int* nums, int numsSize){
int i, thisSum, maxSum;
thisSum = 0;
maxSum = 0x80000000;
for (i = 0; i < numsSize; i++) {
thisSum += nums[i];
if (thisSum > maxSum)
maxSum = thisSum;
if (thisSum < 0)
thisSum = 0;
}
return maxSum;
}
