<leetcode系列> Contains Duplicate II

Contains Duplicate II

Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k.

题意如下:

给定整数数组nums和整数k, 判断在数组中是否存在nums[i] = nums[j], 并且i, j满足i, j之差不大于k.

和Contains Duplicate I 和类似之处.代码中也用到了Contains Duplicate I的代码.

题目链接: https://leetcode.com/problems/contains-duplicate-ii/

Contains Duplicate I 链接: https://leetcode.com/problems/contains-duplicate/

由于很少使用标准库中的各种数据结果,如vector, map等,对于string这种类型,其中的各种成员函数都要先百度,才只有有什么可以调用,不同库中的string又有不小的差异.故我选用的是C语言.代码如下:

// 在长度为numsSize的数组nums中,是否包含相同的数.utils是为了避免重复申请工具数组.
bool containsDuplicate(int* nums, int numsSize, int* utils) {
if ((0 >= numsSize) || (NULL == nums)) {
return false;
}

for (int i = 0; i < numsSize; ++i) {
++utils[nums[i]];
if (utils[nums[i]] > 1) {
return true;
}
}

return false;
}

bool containsNearbyDuplicate(int* nums, int numsSize, int k) {
if ((NULL == nums) || (1 >= numsSize) || (0 >= k)) {
return false;
}

int maxVal = -2147483648;
int minVal = 2147483647;
for (int i = 0; i < numsSize; ++i) {
if (maxVal < nums[i]) {
maxVal = nums[i];
}
if (minVal > nums[i]) {
minVal = nums[i];
}
}
for (int i = 0; i < numsSize; ++i) {
nums[i] -= minVal;
}
maxVal -= minVal;
if (maxVal == 0) { // 正常情况下,maxVal应该是大于等于0的.如果等于零, 则所有数相等
return true;
}
// 索引到下标maxVal, 则需要maxVal + 1长度的数组
int* utils = (int*) malloc((maxVal + 1) * sizeof(int));
memset(utils, '\0', (maxVal + 1) * sizeof(int));

bool ret = false;
if (k >= numsSize) {
ret = containsDuplicate(nums, numsSize, utils);
} else {
for (int i = 0; i < numsSize - k; ++i) {
if (containsDuplicate(nums + i, k + 1, utils)) {
ret = true;
break;
}
memset(utils, '\0', (maxVal + 1) * sizeof(int));
}
}

free(utils);
utils = NULL;
return ret;
}