@@ -515,7 +515,8 @@ impl Solution {
515515```
516516
517517** C:**
518- ``` c(版本一)
518+ ```
519+ //C (版本一) 左闭右闭区间 [left, right]
519520int search(int* nums, int numsSize, int target){
520521 int left = 0;
521522 int right = numsSize-1;
@@ -542,21 +543,21 @@ int search(int* nums, int numsSize, int target){
542543}
543544```
544545```
545- C (版本二)
546+ C (版本二) 左闭右开区间 [left, right)
546547int search(int* nums, int numsSize, int target){
547548 int length = numsSize;
548549 int left = 0;
549- int right = length;//定义target在左闭右开的区间里,即:[left, right)
550+ int right = length;//定义target在左闭右开的区间里,即:[left, right)
550551 int middle = 0;
551- while(left < right){// left == right时,区间[left, right)属于空集,所以用 < 避免该情况
552+ while(left < right){// left == right时,区间[left, right)属于空集,所以用 < 避免该情况
552553 int middle = left + (right - left) / 2;
553554 if(nums[middle] < target){
554555 //target位于(middle , right) 中为保证集合区间的左闭右开性,可等价为[middle + 1,right)
555556 left = middle + 1;
556557 }else if(nums[middle] > target){
557558 //target位于[left, middle)中
558559 right = middle ;
559- }else{// nums[middle] == target ,找到目标值target
560+ }else{// nums[middle] == target ,找到目标值target
560561 return middle;
561562 }
562563 }
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