题目：

题解：

int maximumGap(int* nums, int numsSize) {if (numsSize < 2) {return 0;}int maxVal = INT_MIN, minVal = INT_MAX;for (int i = 0; i < numsSize; ++i) {maxVal = fmax(maxVal, nums[i]);minVal = fmin(minVal, nums[i]);}int d = fmax(1, (maxVal - minVal) / (numsSize - 1));int bucketSize = (maxVal - minVal) / d + 1;int bucket[bucketSize][2];memset(bucket, -1, sizeof(bucket));  // 存储 (桶内最小值，桶内最大值) 对，(-1, -1) 表示该桶是空的for (int i = 0; i < numsSize; i++) {int idx = (nums[i] - minVal) / d;if (bucket[idx][0] == -1) {bucket[idx][0] = bucket[idx][1] = nums[i];} else {bucket[idx][0] = fmin(bucket[idx][0], nums[i]);bucket[idx][1] = fmax(bucket[idx][1], nums[i]);}}int ret = 0;int prev = -1;for (int i = 0; i < bucketSize; i++) {if (bucket[i][0] == -1) continue;if (prev != -1) {ret = fmax(ret, bucket[i][0] - bucket[prev][1]);}prev = i;}return ret;
}