网站的需求分析包括哪些下载官方正版百度

https://leetcode.cn/problems/maximum-product-of-two-elements-in-an-array/description/
第一种：常规解法，遍历两次数组根据条件比较出最大的即可

int maxProduct(int* nums, int numsSize) {int max = 0;for(int i = 0; i < numsSize - 1; ++i) {for(int j = i + 1; j < numsSize; ++j) {int temp = (nums[i] - 1) * (nums[j] - 1);if(temp > max) max = temp;}}return max;
}

第二种：使用排序，先排序数组，然后直接将最大的和次大的做运算得出结果。

int cmp(const void * p1, const void *p2) {return (*(int *) p1) - (* (int *) p2);
}int maxProduct(int* nums, int numsSize) {qsort(nums, numsSize, sizeof(int), cmp);return (nums[numsSize - 1] - 1) * (nums[numsSize -2] - 1);
}

https://leetcode.cn/problems/max-consecutive-ones/description/
遍历数组，将1全部加起来，出现0就重置。

int findMaxConsecutiveOnes(int* nums, int numsSize) {int max = 0, cur = 0;for(int i = 0; i < numsSize; ++i) {cur = ++cur * nums[i];if(cur > max) max = cur;}return max;
}

这一个和上面是一样的思路，只是实现不同而已

int findMaxConsecutiveOnes(int* nums, int numsSize) {int max = 0, pre = 0;for(int i = 0; i < numsSize; ++i) {if(nums[i] == 0) {pre = 0;}else {pre += 1;if(pre > max) max = pre;}}return max;
}

https://leetcode.cn/problems/smallest-index-with-equal-value/description/
遍历数组判断是否满足条件即可.

int smallestEqual(int* nums, int numsSize) {for(int i = 0; i < numsSize; ++i) {if(i % 10 == nums[i]) return i;}return -1;
}

https://leetcode.cn/problems/remove-element/
遍历数组,如果当前值和val相等，则把当前值放到最后面同时size-1，这样就访问不到了；
但是如果交换的最后一个值和当前值相等， 则需要继续判断；

int removeElement(int* nums, int numsSize, int val) {for(int i = 0; i < numsSize; ++i) {while(i < numsSize && nums[i] == val) {int temp = nums[i];nums[i] = nums[numsSize - 1];nums[numsSize - 1] = temp;--numsSize;}}return numsSize;
}

https://leetcode.cn/problems/non-decreasing-array/description/
第一遍错误做法：

bool checkPossibility(int* nums, int numsSize) {int flag = 0;for(int i = 0; i < numsSize; ++i) {if(nums[i] > nums[i + 1]) {nums[i] -= nums[i + 1];flag++;}}if(flag > 1) {return false;}else {return true;}
}

需要多加写条件判断，还是太年轻了~

bool checkPossibility(int* nums, int numsSize) {int flag = 0;int pos = -1;for(int i = 0; i < numsSize - 1; ++i) {if(nums[i] > nums[i + 1]) {pos = i;flag++;}}if(flag >= 2) return false;if(flag == 0) return true;if(pos == 0 || nums[pos - 1] <= nums[pos + 1]) return true;if(pos == numsSize - 2 || nums[pos] <= nums[pos + 2]) return true;return false;
}