数组类36--在排序数组中查找元素的第一个和最后一个位置（M）

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AC代码

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] ans = new int[2];
        ans[0] = -1;
        ans[1] = -1;
        if(nums == null || nums.length == 0) {
            return ans;
        }
        int outLeft = 0;
        int outRight = nums.length - 1;
        while(outLeft <= outRight) {
            int mid = outLeft + (outRight - outLeft) / 2;
            if(nums[mid] == target) {
                ans[0] = findLeft(nums, target, outLeft, mid);
                ans[1] = findRight(nums, target, mid, outRight);
                return ans;
            }else if(nums[mid] > target) {
                outRight = mid - 1;
            }else {
                outLeft = mid + 1;
            }

        }
        return ans;
    }
    private int findLeft(int[] nums, int target, int left, int right) {
        int ans = right;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if(nums[mid] == target) {
                ans = mid;
                right = mid - 1;
            }else if(nums[mid] < target){
                left = mid + 1;
            }
        }
        return ans;
    }

    private int findRight(int[] nums, int target, int left, int right) {
        int ans = left;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if(nums[mid] == target) {
                ans = mid;
                left = mid + 1;
            }else if(nums[mid] > target){
                right = mid - 1;
            }
        }
        return ans;
    }
}

精髓
AC代码看起来比较复杂，逻辑比较清楚
首先普通二分查找，找到第一个target，如果找不到就是没有，就直接返回
然后分别向左和向右查找区间左端点和右端点，都是基于二分的思想