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| /**
* 这道题要用二分法找到左边的开始位置，以及右边的结束位置
* 同时需要考虑，target不在数组中
* <p>
* 情况一：target 在数组范围的右边或者左边，例如数组{3, 4, 5}，target为2或者数组{3, 4, 5},target为6，此时应该返回{-1, -1}
* 情况二：target 在数组范围中，且数组中不存在target，例如数组{3,6,7},target为5，此时应该返回{-1, -1}
* 情况三：target 在数组范围中，且数组中存在target，例如数组{3,6,7},target为6，此时应该返回{1, 1}
*/
public int[] searchRange(int[] nums, int target) {
if (nums.length == 0) {
return new int[]{-1, -1};
}
int rithtIndex = rithtIndex(nums, target);
int leftIndex = leftIndex(nums, target);
if (rithtIndex == -2 || leftIndex == -2) {
return new int[]{-1, -1};
}
if (rithtIndex - leftIndex > 1) return new int[]{leftIndex + 1, rithtIndex - 1};
return new int[]{-1, -1};
}
public int rithtIndex(int[] nums, int target) {
int l = 0;
int r = nums.length - 1;
int rithtIndex = -2;
while (l <= r) {
int mid = l + (r - l) / 2;
if (nums[mid] > target) {//一直缩小右边界
r = mid - 1;
} else {
l = mid + 1;
rithtIndex = l;
}
}
return rithtIndex;
}
public int leftIndex(int[] nums, int target) {
int l = 0;
int r = nums.length - 1;
int leftIndex = -2;
while (l <= r) {
int mid = l + (r - l) / 2;
if (nums[mid] < target) {//一直缩小右边界
l = mid + 1;
} else {
r = mid - 1;
leftIndex = r;
}
}
return leftIndex;
}
|