34. Search for a Range
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Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1]
.
For example,
Given [5, 7, 7, 8, 8, 10]
and target value 8,
return [3, 4]
.
題意:
根據(jù)給定的排序的整數(shù)數(shù)組和給定值,查找給定值的起始位置和終止位置。如果查找不到指定值則返回起始位置和終止位置都為-1。
處理:
1)在給定的數(shù)組中查找給定值,如果給定值存在,則返回第一次出現(xiàn)的位置;否則返回-1.
2)若返回-1,則返回起始和終止都為-1的數(shù)組;否則,分別前向和后向查找起始和終止位置,返回起始終止位置數(shù)組。
查找指定值使用遞歸進行查找。
1)如果下標中間值和起始下標和終止下標一致,且當前值不是指定值,則返回-1.
2)如果找到指定值則返回下標,否則,返回-1
int findIndex( int *nums, int begin, int end, int target) { int low = begin; int high = end; int mid = ( low + high ) / 2; int value = -1; /* 5 7 7 8 8 10*/ if ( mid == high && low == mid && *(nums + mid) != target ) { return -1; } if ( *( nums + mid ) < target ) { value = findIndex( nums, mid + 1, high, target ); } else if ( *( nums + mid ) > target ) { value = findIndex( nums, low, mid, target ); } else if ( *( nums + mid ) == target ) { return mid; } return value; } /** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */ int* searchRange(int* nums, int numsSize, int target, int* returnSize) { int *dest = NULL; dest = (int *)malloc(sizeof(int) * 3); if ( !dest ) { return NULL; } int mid = findIndex( nums, 0, numsSize - 1, target ); if ( mid == -1 ) { dest[0] = -1; dest[1] = -1; dest[2] = '\0'; *returnSize = 2; return dest; } int cnt = mid; while ( cnt >= 0 && *( nums + cnt ) == target ) { cnt -= 1; } int index = mid; while ( index < numsSize && *( nums + index ) == target ) { index += 1; } dest[0] = cnt + 1; dest[1] = index - 1; dest[2] = '\0'; *returnSize = 2; return dest; }