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前言

使用左閉右閉區(qū)間的二分查找時， 最后low一定是被查找元素的插入位置，若查找的數(shù)帶小數(shù)，low-1， 便是最終結(jié)果

一、704. 二分查找

1、左閉右閉

class Solution {
    public int search(int[] nums, int target) {
        int low = 0, high = nums.length-1, mid = 0;
        while(low <= high){
            mid = (low + high)/2;
            if(nums[mid] == target){
                return mid;
            }else if(nums[mid] < target){
                low = mid + 1;
            }else{
                high = mid - 1;
            }
        }
        return -1;
    }
}

2、左閉右開文章來源地址http://www.zghlxwxcb.cn/news/detail-657723.html

class Solution {
    public int search(int[] nums, int target) {
        int low = 0, high = nums.length, mid = 0;
        while(low < high){
            mid = (low + high)/2;
            if(nums[mid] == target){
                return mid;
            }else if(nums[mid] < target){
                low = mid + 1;
            }else{
                high = mid;
            }
        }
        return -1;
    }
}

二、35. 搜索插入位置

class Solution {
    public int searchInsert(int[] nums, int target) {
        int low = 0, high = nums.length, mid;
        while(low < high){
            mid = (low + high)/2;
            if(nums[mid] == target){
                return mid;
            }else if(nums[mid] < target){
                low = mid + 1;
            }else{
                high = mid;
            }
        }
        return low;
    }
}

三、34. 在排序數(shù)組中查找元素的第一個和最后一個位置

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int low = 0, high = nums.length, mid; int[] res = {-1, -1};
        while(low < high){
            mid = (low + high)/2;
            if(nums[mid] == target){
                res[0] = res[1] = mid;
                while(res[0]-1 >= 0 && nums[res[0] - 1] == target){
                    res[0] -= 1;
                }
                while(res[1] + 1 < nums.length && nums[res[1] + 1] == target){
                    res[1] += 1;
                }
                return res;
            }else if(nums[mid] < target){
                low = mid + 1;
            }else{
                high = mid;
            }
        }
        return res;
    }
}

四、69. x 的平方根

class Solution {
    public int mySqrt(int x) {
        int low = 0, high = x, mid;
        if(x == 0 || x == 1){
            return x;
        }
        while(low <= high){
            mid = (low + high)/2;
            if(x / mid == mid){
                return mid;
            }else if(x / mid > mid){
                low = mid +1;
            }else{
                high = mid -1;
            }
        }
        return low - 1;
    }
}

五、367. 有效的完全平方數(shù)

lass Solution {
    public boolean isPerfectSquare(int num) {
        int x = 1;
        while(num > 0){
            num -= x;
            x += 2;
        }
        return num == 0;
    }
}

六、27. 移除元素

class Solution {
    public int removeElement(int[] nums, int val) {
        int i = 0, j = 0;
        for(;i < nums.length; ){
            if(nums[i] != val){
                nums[j] = nums[i];
                i ++; j ++;
            }else{
                i ++;
            }
        }
        return j;
    }
}

七、26. 刪除有序數(shù)組中的重復(fù)項(xiàng)

class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums.length == 1){
            return 1;
        }
        int i = 1, j = 0;
        for(; i < nums.length; ){
            if(nums[i] != nums[j]){
                nums[++j] = nums[i++];
            }else{
                i ++;
            }
        }
        return j + 1;
    }
}

八、283. 移動零

class Solution {
    public void moveZeroes(int[] nums) {
        int i = 0, j = 0, len = nums.length;
        if(len == 1)return;
        while(i < len){
            if(nums[i] != 0){
                nums[j] = nums[i];
                if(i == j){
                    i ++;
                }else{
                    nums[i++] = 0;
                }
                j ++;
            }else{
                i ++;
            }
        }
    }
}

九、844. 比較含退格的字符串

class Solution {
    public boolean backspaceCompare(String s, String t) {
        Deque<Character> deq1 = new ArrayDeque<>();
        Deque<Character> deq2 = new ArrayDeque<>();
        char[] ch1 = s.toCharArray();
        char[] ch2 = t.toCharArray();
        for(int i = 0; i < ch1.length; i ++){
            if(ch1[i] != '#'){
                deq1.offerFirst(ch1[i]);
            }else if(!deq1.isEmpty()){
                deq1.pollFirst();
            }
        }
        for(int j = 0; j < ch2.length; j ++){
            if(ch2[j] != '#'){
                deq2.offerFirst(ch2[j]);
            }else if(!deq2.isEmpty()){
                deq2.pollFirst();
            }
        }
        while(!deq1.isEmpty() && !deq2.isEmpty()){
            char c1 = deq1.pollFirst();
            char c2 = deq2.pollFirst();
            if(c1 != c2){
                return false;
            }
        }
        return deq1.isEmpty() && deq2.isEmpty();
    }
}

十、977. 有序數(shù)組的平方

class Solution {
    public int[] sortedSquares(int[] nums) {
        Deque<Integer> deq1 = new LinkedList<>();
        Deque<Integer> deq2 = new LinkedList<>();
        for(int i = 0; i < nums.length; i ++){
            if(nums[i] <= 0){
                deq1.offerLast(nums[i] * nums[i]);
            }else{
                deq2.offerLast(nums[i] * nums[i]);
            }
        }
        int k = 0;
        while(!deq1.isEmpty() && !deq2.isEmpty()){
            if(deq1.peekLast() <= deq2.peekFirst()){
                nums[k ++] = deq1.pollLast();
            }else{
                nums[k ++] = deq2.pollFirst();
            }
        }
        while(!deq1.isEmpty()){
            nums[k ++] = deq1.pollLast();
        }
        while(!deq2.isEmpty()){
            nums[k ++] = deq2.pollFirst();
        }
        return nums;
    }
}

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